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df2ad2e07c086bbf189df37beea16795

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_1

Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ $a_2$ $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$

One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of $a_1$ , then use that to calculate $a_2$ and sum another arithmetic series to get our answer. A somewhat quicker method is to do the following: for each $n \geq 1$ , we have $a_{2n - 1} = a_{2n} - 1$ . We can substitute this into our given equation to get $(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137$ . The left-hand side of this equation is simply $2(a_2 + a_4 + \ldots + a_{98}) - 49$ , so our desired value is $\frac{137 + 49}{2} = \boxed{093}$

093

df2ad2e07c086bbf189df37beea16795

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_1

Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ $a_2$ $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$

We want to find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ , which can be rewritten as $a_1+1+a_2+2+a_3+\ldots+a_{49}+49 \implies a_1+a_2+a_3+\ldots+a_{49}+\frac{49 \cdot 50}{2}$ . We can split $a_1+a_2+a_3+\ldots+a_{98}$ into two parts: \[a_1+a_2+a_3+\ldots+a_{49}\] and \[a_{50}+a_{51}+a_{52}+\ldots+a_{98}\] Note that each term in the second expression is $49$ greater than the corresponding term, so, letting the first equation be equal to $x$ , we get $a_1+a_2+a_3+\ldots+a_{98}=137=2x+49^2 \implies x=\frac{137-49^2}{2}$ . Calculating $49^2$ by sheer multiplication is not difficult, but you can also do $(50-1)(50-1)=2500-100+1=2401$ . We want to find the value of $x+\frac{49 \cdot 50}{2}=x+49 \cdot 25=x+1225$ . Since $x=\frac{137-2401}{2}$ , we find $x=-1132$ $-1132+1225=\boxed{93}$

93

df2ad2e07c086bbf189df37beea16795

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_1

Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ $a_2$ $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$

Since we are dealing with an arithmetic sequence, \[a_2+a_4+a_6+a_8+\ldots+a_{98} = 49a_{50}\] We can also figure out that \[a_1+a_2+a_3+\ldots+a_{98} = a_1 + 97a_{50} = 137\] \[a_1 = a_{50}-49 \Rightarrow 98a_{50}-49 = 137\] Thus, $49a_{50} = \frac{137 + 49}{2} = \boxed{093}$

093

022c8702785e13f53c1d47c6832bd2b1

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_2

The integer $n$ is the smallest positive multiple of $15$ such that every digit of $n$ is either $8$ or $0$ . Compute $\frac{n}{15}$

Any multiple of 15 is a multiple of 5 and a multiple of 3. Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0. The sum of the digits of any multiple of 3 must be divisible by 3. If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$ . For this number to be divisible by 3, $a$ must be divisible by 3. We also know that $a>0$ since $n$ is positive. Thus $n$ must have at least three copies of the digit 8. The smallest number which meets these two requirements is 8880. Thus the answer is $\frac{8880}{15} = \boxed{592}$

592

911ac42e746502c1beb963fb72b27c1d

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_3

point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$

By the transversals that go through $P$ , all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = \dfrac{ab\sin C}{2}$ to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as $2x,\ 3x,\ 7x$ . Thus, the corresponding side on the large triangle is $12x$ , and the area of the triangle is $12^2 = \boxed{144}$

144

911ac42e746502c1beb963fb72b27c1d

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_3

point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$

Alternatively, since the triangles are similar by $AA$ , then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume $\dfrac{base}{height} = \dfrac{2}{1}.$ That means that the base of $t_{1}$ is 4, the base of $t_{2}$ is 6, and the base of $t_{3}$ is 14. Since the quadrilaterals underneath $t_{1}$ and $t_{2}$ are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is $4 + 14 + 6 = 24$ . Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is $\dfrac{1}{2} \cdot 24 \cdot 12 = \boxed{144}$

144

911ac42e746502c1beb963fb72b27c1d

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_3

point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$

The base of $\triangle{ABC}$ is $BC$ . Let the base of $t_1$ be $x$ , the base of $t_2$ be $y$ , and the base of $t_3$ be $z$ . Since $\triangle{ABC}, t_1, t_2,$ and $t_3$ are all similar, the sections in $\triangle{ABC}$ that aren't $t_1,t_2,$ or $t_3$ are all parallelograms. Hence, $BC=x+z+y$ . We can relate $t_1,t_2,$ and $t_3$ by the square root of the ratio of their areas. $\sqrt{\frac{4}{9}}=\frac{2}{3}$ and $\sqrt{\frac{4}{49}}=\frac{2}{7}$ so $y=\frac{3x}{2}$ and $z=\frac{7x}{2}$ $x+\frac{7x}{2}+\frac{3x}{2}=6x$ , so $\triangle{ABC}$ has a base that is $6$ times $t_1$ $[\triangle{ABC}]=36[t_1]=36 \cdot 4=\boxed{144}$

144

911ac42e746502c1beb963fb72b27c1d

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_3

point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$

Since the three lines through $P$ are parallel to the sides, $t_1$ $t_2$ $t_3$ , and $\triangle{ABC}$ are similar by $AA$ similarity. Suppose the area of $\triangle{ABC}$ is $x^2$ , so the ratio of the base of $t_1$ to the base of $t_2$ to the base of $t_3$ to the base of $\triangle{ABC}$ is $2:3:7:x$ . Because the quadrilaterals below $t_1$ and $t_2$ are parallelograms, the base of $\triangle{ABC}$ is equal to the sum of the bases of $t_1, t_2,$ and $t_3$ . Therefore, $x$ equals $2+3+7=12$ so the area of $\triangle{ABC}$ equals $x^2=12^2=\boxed{144}.$

144

b16cdd62a11dbaed32fad7e978b4ff72

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_4

Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$ . However, if $68$ is removed, the average of the remaining numbers drops to $55$ . What is the largest number that can appear in $S$

Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$ We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$ The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$ s and one $660-11\cdot1=\boxed{649}.$

649

b16cdd62a11dbaed32fad7e978b4ff72

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_4

Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$ . However, if $68$ is removed, the average of the remaining numbers drops to $55$ . What is the largest number that can appear in $S$

Suppose that $S$ has $n$ numbers other than the $68.$ We have the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{Count} & \textbf{Arithmetic Mean} & \textbf{Sum} \\ \hline & & & \\ [-2.5ex] \textbf{Initial} & n+1 & 56 & 56(n+1) \\ \hline & & & \\ [-2.5ex] \textbf{Final} & n & 55 & 55n \end{array}\] We are given that \[56(n+1)-68=55n,\] from which $n=12.$ It follows that the sum of the remaining numbers in $S$ is $55n=660.$ We continue with the last paragraph of Solution 1 to get the answer $\boxed{649}.$

649

6e9fbb19e61930d3dc1aa3d8d3ae6e81

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$

Use the change of base formula to see that $\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5$ ; combine denominators to find that $\frac{\log ab^3}{3\log 2} = 5$ . Doing the same thing with the second equation yields that $\frac{\log a^3b}{3\log 2} = 7$ . This means that $\log ab^3 = 15\log 2 \Longrightarrow ab^3 = 2^{15}$ and that $\log a^3 b = 21\log 2 \Longrightarrow a^3 b = 2^{21}$ . If we multiply the two equations together, we get that $a^4b^4 = 2^{36}$ , so taking the fourth root of that, $ab = 2^9 = \boxed{512}$

512

6e9fbb19e61930d3dc1aa3d8d3ae6e81

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$

We can simplify our expressions by changing everything to a common base and by pulling exponents out of the logarithms. The given equations then become $\frac{\ln a}{\ln 8} + \frac{2 \ln b}{\ln 4} = 5$ and $\frac{\ln b}{\ln 8} + \frac{2 \ln a}{\ln 4} = 7$ . Adding the equations and factoring, we get $(\frac{1}{\ln 8}+\frac{2}{\ln 4})(\ln a+ \ln b)=12$ . Rearranging we see that $\ln ab = \frac{12}{\frac{1}{\ln 8}+\frac{2}{\ln 4}}$ . Again, we pull exponents out of our logarithms to get $\ln ab = \frac{12}{\frac{1}{3 \ln 2} + \frac{2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2$ . This means that $\frac{\ln ab}{\ln 2} = 9$ . The left-hand side can be interpreted as a base-2 logarithm, giving us $ab = 2^9 = \boxed{512}$

512

6e9fbb19e61930d3dc1aa3d8d3ae6e81

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$

This solution is very similar to the above two, but it utilizes the well-known fact that $\log_{m^k}{n^k}= \log_m{n}.$ Thus, $\log_8a+\log_4b^2=5 \Rightarrow \log_{2^3}{(\sqrt[3]{a})^3} + \log_{2^2}{b^2} = 5 \Rightarrow \log_2{\sqrt[3]{a}} + \log_2{b} = 5 \Rightarrow \log_2{\sqrt[3]{a}b} = 5.$ Similarly, $\log_8b+\log_4a^2=7 \Rightarrow \log_2{\sqrt[3]{b}a} = 7.$ Adding these two equations, we have $\log_2{a^{\frac{4}{3}}b^{\frac{4}{3}}} = 12 \Rightarrow ab = 2^{12\times\frac{3}{4}} = 2^9 = \boxed{512}$

512

6e9fbb19e61930d3dc1aa3d8d3ae6e81

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$

We can change everything to a common base, like so: $\log_8{a} + \log_8{b^3} = 5,$ $\log_8{b} + \log_8{a^3} = 7.$ We set the value of $\log_8{a}$ to $x$ , and the value of $\log_8{b}$ to $y.$ Now we have a system of linear equations: \[x + 3y = 5,\] \[y + 3x = 7.\] Now add the two equations together then simplify, we'll get $x+y=3$ . So $\log_8{ab} = \log_8{a} + \log_8{b} = 3$ $ab = 8^3 = \boxed{512}$

512

6e9fbb19e61930d3dc1aa3d8d3ae6e81

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$

Add the two equations to get $\log_8 {a}+ \log_8 {b}+ \log_{a^2}+\log_{b^2}=12$ . This can be simplified with the log property $\log_n {x}+\log_n {y}=log_n {xy}$ . Using this, we get $\log_8 {ab}+ \log_4 {a^2b^2}=12$ . Now let $\log_8 {ab}=c$ and $\log_4 {a^2b^2}=k$ . Converting to exponents, we get $8^c=ab$ and $4^k=(ab)^2$ . Sub in the $8^c$ to get $k=3c$ . So now we have that $k+c=12$ and $k=3c$ which gives $c=3$ $k=9$ . This means $\log_4 {a^2b^2}=9$ so $4^9=(ab)^2 \implies ab=(2^2)^9 \implies 2^9 \implies \boxed{512}$

512

6e9fbb19e61930d3dc1aa3d8d3ae6e81

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$

By properties of logarithms, we know that $\log_8 {a}+ \log_4 {b ^ 2} = \log_2 {a ^ {1/3}}+ \log_2 {b} = 5$ Using the fact that $\log_a {b} + \log_a {c} = log_a{b*c}$ , we get $\log_2 {a^{1/3} * b} = 5$ Similarly, we know that $\log_2 {a * b^{1/3}} = 7$ From these two equations, we get $a^{1/3} * b = 2^5$ and $a * b^{1/3} = 2^7$ Multiply the two equations to get $a^{4/3} * b^{4/3} = 2^{12}$ . Solving, we get that $a*b = 2^{12*3/4} = 2^9 =$ $\boxed{512}$

512

6e9fbb19e61930d3dc1aa3d8d3ae6e81

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$

Adding both of the equations, we get \[\log_8{ab} +2\log_4{ab}=12\] Furthermore, we see that $\log_4 {ab}$ is $\frac{3}{2}$ times $\log_8 {ab}.$ Substituting $\log_8 {ab}$ as $x,$ we get $x+3x=12,$ so $x=3.$ Therefore, we have $\log_8 {ab} = 3,$ so $ab= 8^3=\boxed{512}$ ~ math_comb01

512

b7825445ca1db3615e3d427fd238e4c6

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_6

Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?

1984 AIME-6.png The line passes through the center of the bottom circle; hence it is the circle's diameter and splits the circle into two equal areas. For the rest of the problem, we do not have to worry about that circle. Draw the midpoint of $\overline{AC}$ (the centers of the other two circles), and call it $M$ . If we draw the feet of the perpendiculars from $A,C$ to the line (call $E,F$ ), we see that $\triangle AEM\cong \triangle CFM$ by HA congruency ; hence $M$ lies on the line. The coordinates of $M$ are $\left(\frac{19+14}{2},\frac{84+92}{2}\right) = \left(\frac{33}{2},88\right)$ Thus, the slope of the line is $\frac{88 - 76}{\frac{33}{2} - 17} = -24$ , and the answer is $\boxed{024}$

024

b7825445ca1db3615e3d427fd238e4c6

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_6

Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?

First of all, we can translate everything downwards by $76$ and to the left by $14$ . Then, note that a line passing through a given point intersecting a circle with a center as that given point will always cut the circle in half, so we can re-phrase the problem: Two circles, each of radius $3$ , are drawn with centers at $(0, 16)$ , and $(5, 8)$ . A line passing through $(3,0)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line? Note that this is equivalent to finding a line such that the distance from $(0,16)$ to the line is the same as the distance from $(5,8)$ to the line. Let the line be $y - ax - b = 0$ . Then, we have that: \[\frac{|-5a + 8 - b|}{\sqrt{a^2+1}}= \frac{|16 - b|}{\sqrt{a^2+1}} \Longleftrightarrow |-5a+8-b| = |16-b|\] We can split this into two cases. Case 1: $16-b = -5a + 8 - b \Longleftrightarrow a = -\frac{8}{5}$ In this case, the absolute value of the slope of the line won’t be an integer, and since this is an AIME problem, we know it’s not possible. Case 2: $b-16 = -5a + 8 - b \Longleftrightarrow 2b + 5a = 24$ But we also know that it passes through the point $(3,0)$ , so $-3a-b = 0 \Longleftrightarrow b = -3a$ . Plugging this in, we see that $2b + 5a = 24 \Longleftrightarrow a = -24$ $\boxed{24}$

24

b7825445ca1db3615e3d427fd238e4c6

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_6

Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?

Consider the region of the plane between $x=16$ and $x=17$ . The parts of the circles centered at $(14,92)$ and $(19,84)$ in this region have equal area. This is by symmetry- the lines defining the region are 2 units away from the centers of each circle and therefore cut off congruent segments. We will draw the line in a way that uses this symmetry and makes identical cuts on the circles. Since $(17,76)$ is $8$ units below the center of the lower circle, we will have the line exit the region $8$ units above the center of the upper circle, at $(16,100)$ . We then find that the slope of the line is $-24$ and our answer is $\boxed{024}$

024

b7825445ca1db3615e3d427fd238e4c6

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_6

Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?

We can redefine the coordinate system so that the center of the center circle is the origin, for easier calculations, as the slope of the line and the congruence of the circles do not depend on it. $O_1=(-3, 16)$ $O_2=(0,0)$ , and $O_3=(2,8)$ . A line bisects a circle iff it passes through the center. Therefore, we can ignore the bottom circle because it contributes an equal area with any line. A line passing through the centroid of any plane system with two perpendicular lines of reflectional symmetry bisects it. We have defined two points of the line, which are $(0,0)$ and $(-\frac{1}{2},12)$ . We use the slope formula to calculate the slope, which is $-24$ , leading to an answer of $\boxed{024}$ $QED \blacksquare$

024

b7825445ca1db3615e3d427fd238e4c6

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_6

Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?

Notice that any line that passes through the bottom circle's center cuts it in half, so all we really care about are the top two circles. Suppose $\ell$ is the desired line. Draw lines $\ell_1$ and $\ell_2$ both parallel to $\ell$ such that $\ell_1$ passes through $(14,92)$ and $\ell_2$ passes through $(19,84)$ . Clearly, $\ell$ must be the "average" of $\ell_1$ and $\ell_2$ . Suppose $\ell:=y=mx+b, \ell_1:=y=mx+c, \ell_2:=y=mx+d$ . Then $b=76-17m, c=92-14m, d=84-19m$ . So we have that \[76-17m=\frac{92-14m+84-19m}{2},\] which yields $m=-24$ for an answer of $\boxed{024}$

024

a991b0df9bf6288b350d4c4dc326105f

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_7

The function f is defined on the set of integers and satisfies $f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}$ Find $f(84)$

Define $f^{h} = f(f(\cdots f(f(x))\cdots))$ , where the function $f$ is performed $h$ times. We find that $f(84) = f(f(89)) = f^2(89) = f^3(94) = \ldots f^{y}(1004)$ $1004 = 84 + 5(y - 1) \Longrightarrow y = 185$ . So we now need to reduce $f^{185}(1004)$ Let’s write out a couple more iterations of this function: \begin{align*}f^{185}(1004)&=f^{184}(1001)=f^{183}(998)=f^{184}(1003)=f^{183}(1000)\\ &=f^{182}(997)=f^{183}(1002)=f^{182}(999)=f^{183}(1004)\end{align*} So this function reiterates with a period of 2 for $x$ . It might be tempting at first to assume that $f(1004) = 1001$ is the answer; however, that is not true since the solution occurs slightly before that. Start at $f^3(1004)$ \[f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}\]

997

a991b0df9bf6288b350d4c4dc326105f

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_7

The function f is defined on the set of integers and satisfies $f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}$ Find $f(84)$

Assume that $f(84)$ is to be performed $n+1$ times. Then we have \[f(84)=f^{n+1}(84)=f(f^n(84+5))\] In order to find $f(84)$ , we want to know the smallest value of \[f^n(84+5)\ge1000\] Because then \[f(84)=f(f^n(84+5))=(f^n(84+5))-3\] From which we'll get a numerical value for $f(84)$ Notice that the value of $n$ we expect to find is basically the smallest $n$ such that after $f(x)=f(f(x+5))$ is performed $\frac{n}{2}$ times and then $f(x)=x-3$ is performed back $\frac{n}{2}$ times, the result is greater than or equal to $1000$ In this case, the value of $n$ for $f(84)$ is $916$ , because \[84+\frac{916}{2}\cdot5-\frac{916}{2}\cdot3=1000\Longrightarrow f^{916}(84+5))=1000\] Thus \[f(84)=f(f^{916}(84+5))=f(1000)=1000-3=\boxed{997}\]

997

da177a2184c5c4496034a7ae79465562

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_8

The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane . Determine the degree measure of $\theta$

We shall introduce another factor to make the equation easier to solve. If $r$ is a root of $z^6+z^3+1$ , then $0=(r^3-1)(r^6+r^3+1)=r^9-1$ . The polynomial $x^9-1$ has all of its roots with absolute value $1$ and argument of the form $40m^\circ$ for integer $m$ (the ninth degree roots of unity ). Now we simply need to find the root within the desired range that satisfies our original equation $x^6 + x^3 + 1 = 0$ This reduces $\theta$ to either $120^{\circ}$ or $160^{\circ}$ . But $\theta$ can't be $120^{\circ}$ because if $r=\cos 120^\circ +i\sin 120^\circ$ , then $r^6+r^3+1=3$ . (When we multiplied by $r^3 - 1$ at the beginning, we introduced some extraneous solutions, and the solution with $120^\circ$ was one of them.) This leaves $\boxed{160}$

160

da177a2184c5c4496034a7ae79465562

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_8

The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane . Determine the degree measure of $\theta$

The substitution $y=z^3$ simplifies the equation to $y^2+y+1 = 0$ . Applying the quadratic formula gives roots $y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$ , which have arguments of $120$ and $240,$ respectively. We can write them as $z^3 = \cos 240^\circ + i\sin 240^\circ$ and $z^3 = \cos 120^\circ + i\sin 120^\circ$ . So we can use De Moivre's theorem (which I would suggest looking at if you never heard of it before) to find the fractional roots of the expressions above! For $\cos 240^\circ + i\sin 240$ we have $(\cos 240^\circ + i\sin 240^\circ)^{1/3}$ $\Rightarrow$ $\cos 80^\circ + i\sin 80^\circ, \cos 200^\circ + i\sin200^\circ,$ and $\cos 320^\circ + i\sin320^\circ.$ Similarly for $(\cos 120^\circ + i\sin 120^\circ)^{1/3}$ , we have $\cos 40^\circ + i\sin 40^\circ, \cos 160^\circ + i\sin 160^\circ,$ and $\cos 280^\circ + i\sin 280^\circ.$ The only argument out of all these roots that fits the description is $\theta = \boxed{160}$

160

1decc201a8ac6f680ca8c32740b8a5a2

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_9

In tetrahedron $ABCD$ edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$ . These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$

Position face $ABC$ on the bottom. Since $[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}$ , we find that $h_{ABD} = 8$ . Because the problem does not specify, we may assume both $ABC$ and $ABD$ to be isosceles triangles. Thus, the height of $ABD$ forms a $30-60-90$ with the height of the tetrahedron. So, $h = \frac{1}{2} (8) = 4$ . The volume of the tetrahedron is thus $\frac{1}{3}Bh = \frac{1}{3} \cdot15 \cdot 4 = \boxed{020}$

020

1decc201a8ac6f680ca8c32740b8a5a2

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_9

In tetrahedron $ABCD$ edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$ . These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$

It is clear that $DX=8$ and $CX=10$ where $X$ is the foot of the perpendicular from $D$ and $C$ to side $AB$ . Thus $[DXC]=\frac{ab\sin{c}}{2}=20=5 \cdot h \rightarrow h = 4$ where h is the height of the tetrahedron from $D$ . Hence, the volume of the tetrahedron is $\frac{bh}{3}=15\cdot \frac{4}{3}=\boxed{020}$ ~ Mathommill

020

1decc201a8ac6f680ca8c32740b8a5a2

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_9

In tetrahedron $ABCD$ edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$ . These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$

We can use 3D coordinates. Let $A = (0, 0, 0)$ and $B = (3, 0, 0).$ WLOG, let $D = \left(\frac{3}{2}, 8, 0\right)$ , because the area of $\Delta{ABD} = 12$ and the tetrahedron area won't change if we put it somewhere else with $y=8.$ To find $C$ , we can again let the $x$ -coordinate be $\frac{3}{2}$ for simplicity. Note that $C$ is $10$ units away from $AB$ because the area of $\Delta{ABC}$ is $15$ . Since the angle between $ABD$ and $ABC$ is $30^\circ$ , we can form a 30-60-90 triangle between $A$ $B$ , and an altitude dropped from $C$ onto face $ABD$ . Since $10$ is the hypotenuse, we get $5\sqrt{3}$ and $5$ as legs. Then $y=5\sqrt{3}$ and $z=5$ , so $C = \left(\frac{3}{2}, 5\sqrt{3}, 5\right).$ (I highly advise you to draw both the tetrahedron and 30-60-90 triangle to get a better perspective.) Now, we can move onto vectors. To find the volume of the tetrahedron, we use the formula $\frac{1}{3}Bh.$ Letting $\Delta{ABC}$ be the base we have $B = 15$ (from the problem statement). We need to find the distance between $D$ and $ABC$ , and to do this, we should find the projection of $D$ onto face $ABC$ Note that we can simplify this to projecting $D$ onto $\mathbf{\overrightarrow{C}}.$ This is because we know the projection will have the same $x$ -coordinate as $D$ and $C$ , as both are $\frac{3}{2}.$ Now we find $\text{proj}_{\mathbf{\overrightarrow{D}}} \mathbf{\overrightarrow{C}}$ , or plugging in our coordinates, $\text{proj}_{\langle\frac{3}{2}, 5\sqrt{3}, 5\rangle} \left\langle\frac{3}{2}, 8, 0\right\rangle$ Let the $x$ -coordinates for both be $0$ for simplicity, because we can always add a $\frac{3}{2}$ at the end. Using the projection formula, we get \[\langle 0, 6, 2\sqrt{3}\rangle.\] Finally, we calculate the distance between $\left(\frac{3}{2}, 6, 2\sqrt{3}\right)$ and $D$ to be $4$ . So the height is $4$ , and plugging into our tetrahedron formula we get \[\frac{1}{3}\cdot 15\cdot 4 = \boxed{20}.\]

20

e6740a52784f53941e2b4f75a1f3d80d

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_10

Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $30$ multiple choice problems and that one's score, $s$ , is computed by the formula $s=30+4c-w$ , where $c$ is the number of correct answers and $w$ is the number of wrong answers. (Students are not penalized for problems left unanswered.)

Let Mary's score, number correct, and number wrong be $s,c,w$ respectively. Then \begin{align*} s&=30+4c-w \\ &=30+4(c-1)-(w-4) \\ &=30+4(c+1)-(w+4). \end{align*} Therefore, Mary could not have left at least five blank; otherwise, one more correct and four more wrong would produce the same score. Similarly, Mary could not have answered at least four wrong (clearly Mary answered at least one right to have a score above $80$ , or even $30$ .) It follows that $c+w\geq 26$ and $w\leq 3$ , so $c\geq 23$ and $s=30+4c-w\geq 30+4(23)-3=119$ . So Mary scored at least $119$ . To see that no result other than $23$ right/ $3$ wrong produces $119$ , note that $s=119\Rightarrow 4c-w=89$ so $w\equiv 3\pmod{4}$ . But if $w=3$ , then $c=23$ , which was the result given; otherwise $w\geq 7$ and $c\geq 24$ , but this implies at least $31$ questions, a contradiction. This makes the minimum score $\boxed{119}$

119

e6740a52784f53941e2b4f75a1f3d80d

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_10

Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $30$ multiple choice problems and that one's score, $s$ , is computed by the formula $s=30+4c-w$ , where $c$ is the number of correct answers and $w$ is the number of wrong answers. (Students are not penalized for problems left unanswered.)

A less technical approach that still gets the job done: Pretend that the question is instead a game, where we are trying to get certain numbers by either adding $4$ or $5.$ The maximum number we can get is $70.$ The goal of the game is to find out what number we can achieve in only ONE method, while all other numbers above that can be achieved with TWO or MORE methods. (Note: This is actually the exact same problem as the original, just reworded differently and now we are adding the score. If this is already confusing, I suggest not looking further.) For example, the number $"21"$ can be achieved with only $1$ method $(4+4+4+4+5).$ However, $25$ , which is a larger number than $21$ , can be achieved with multiple methods (e.g. $5 \cdot 5$ or $4 \cdot 5 + 5$ ), hence $21$ is not the number we are trying to find. If we make a table of adding $4$ or adding $5$ , we will see we get $4, 8, 12, 16, 20,$ etc. if we add only $4$ s and if we add $5$ to those numbers then we will get $9, 13, 17, 21, 25,$ etc. Now a key observation to getting this problem correct is that if we can add one of those previous base numbers to $20$ , then there will be multiple methods (because $20 = 4 \cdot 5 = 5 \cdot 4$ ). Hence, the number we are looking for cannot be $20$ plus one of those base numbers. Instead, it must be $10$ plus that base number, because that results in the same last digit while maintaining only one method to solve. For example, if we start with $4$ , the number $14$ would have only $1$ method to solve, but the number $24$ would have multiple (because $4 + 20 = 24$ and we are trying to avoid adding $20$ ). The largest number we see that is in our base numbers is $21.$ Hence, our maximum number is $21 + 10 = 31.$ Note that if we have the number $25$ , that can be solved via multiple methods, and if we keep repeating our cycle of base numbers, we are basically adding $20$ to a previous base number, which we don't want. And since the maximum number of this game is $31$ , that is the number we subtract from the maximum score of $150$ , so we get $150 - 31 = \boxed{119}.$

119

e6740a52784f53941e2b4f75a1f3d80d

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_10

Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $30$ multiple choice problems and that one's score, $s$ , is computed by the formula $s=30+4c-w$ , where $c$ is the number of correct answers and $w$ is the number of wrong answers. (Students are not penalized for problems left unanswered.)

Based on the value of $c,$ we construct the following table: \[\begin{array}{c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} &\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&&&&&&&&&&&&& \\ [-2.5ex] \boldsymbol{c} &\boldsymbol{\cdots}&\boldsymbol{12}&\boldsymbol{13}&\boldsymbol{14}&\boldsymbol{15}&\boldsymbol{16}&\boldsymbol{17}&\boldsymbol{18}&\boldsymbol{19}&\boldsymbol{20}&\boldsymbol{21}&\boldsymbol{22}&\boldsymbol{23}&\boldsymbol{24}&\boldsymbol{25}&\boldsymbol{26}&\boldsymbol{27}&\boldsymbol{28}&\boldsymbol{29}&\boldsymbol{30} \\ \hline \hline &&&&&&&&&&&&&&&&&&& \\ [-2.25ex] \boldsymbol{s_{\min}} &\cdots&60&65&70&75&80&85&90&95&100&105&110&115&120&125&130&135&140&145&150 \\ \hline &&&&&&&&&&&&&&&&&&& \\ [-2.25ex] \boldsymbol{s_{\max}} &\cdots&78&82&86&90&94&98&102&106&110&114&118&122&126&130&134&138&142&146&150 \end{array}\] For a fixed value of $c,$ note that $s_{\min}$ occurs at $w=30-c,$ and $s_{\max}$ occurs at $w=0.$ Moreover, all integers from $s_{\min}$ through $s_{\max}$ are attainable. To find Mary's score, we look for the lowest score $\boldsymbol{s}$ such that $\boldsymbol{s\geq80}$ and $\boldsymbol{s}$ is contained in exactly one interval. Let $S(c)$ denote the interval of all possible scores $s$ with $c$ correct answers. We need: It follows that the least such value of $c$ is $23,$ from which the lowest such score $s$ is $\boxed{119}.$

119

e6740a52784f53941e2b4f75a1f3d80d

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_10

Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $30$ multiple choice problems and that one's score, $s$ , is computed by the formula $s=30+4c-w$ , where $c$ is the number of correct answers and $w$ is the number of wrong answers. (Students are not penalized for problems left unanswered.)

Given that Mary's score is $30+4c-w$ , two other ways to get that score are $30+4(c+1)-(w+4)$ and $30+4(c-1)-(w-4)$ . Since it is clear that $c>1$ , we must have $w<4$ . In order to minimize the score, assume that $w=3$ . The number of problems left blank must be less than $5$ because of the $30+4(c+1)-(w+4)$ case. In order to minimize the score, assume that the number of problems left blank is $4$ , making the number of correct problems $23$ . Substituting, we get that $s=30+23{\,\cdot\,}4-3$ , so $s=\boxed{119}$

119

09fd4b9b449758113ff8ea1d9eeaf004

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

First notice that there is no difference between the maple trees and the oak trees; we have only two types, birch trees and "non-birch" trees. (If you don't believe this reasoning, think about it. You could also differentiate the tall oak trees from the short oak trees, and the maple trees with many branches as opposed to those with few branches. Indeed, you could keep dividing until you have them each in their own category, but in the end it will not change the probability of the birch trees being near each other. That is, in the end, you multiply the numerator by the number of ways to arrange the oak and maple trees and you also multiply the denominator by the number of ways to arrange the oak and maple trees, making them cancel out.) The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 5 dividers of 8 slots that the birch trees can go in, making ${8\choose5} = 56$ different ways to arrange this. There are ${12 \choose 5} = 792$ total ways to arrange the twelve trees, so the probability is $\frac{56}{792} = \frac{7}{99}$ The answer is $7 + 99 = \boxed{106}$

106

09fd4b9b449758113ff8ea1d9eeaf004

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

Let $b$ $n$ denote birch tree and not-birch tree, respectively. Notice that we only need $4$ $n$ s to separate the $5$ $b$ s. Specifically, \[b,n,b,n,b,n,b,n,b\] Since we have $7$ $n$ s, we are placing the extra $3$ $n$ s into the $6$ intervals beside the $b$ s. Now doing simple casework. If all $3$ $n$ s are in the same interval, there are $6$ ways. If $2$ of the $3$ $n$ s are in the same interval, there are $6\cdot5=30$ ways. If the $n$ s are in $3$ different intervals, there are ${6 \choose 3} =20$ ways. In total there are $6+30+20=56$ ways. There are ${12\choose5}=792$ ways to distribute the birch trees among all $12$ trees. Thus the probability equals $\frac{56}{792}=\frac{7}{99}\Longrightarrow m+n=7+99=\boxed{106}$

106

09fd4b9b449758113ff8ea1d9eeaf004

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

Note that the requested probability is computed by dividing the number of configurations with no adjacent Birch trees by the total number of configurations. We can compute the number of configurations with no adjacent Birch trees using complementary counting and then the Principle of Inclusion-Exclusion. The number of configurations with no adjacent Birch trees is equal to the total number of configurations minus the number of configurations with at least one pair of adjacent Birch trees. The total number of configurations is given by $\frac{12!}{3! \cdot 4! \cdot 5!}$ . To compute the number of configurations with at least one pair of adjacent Birch trees, we use PIE. $\#$ (configurations with at least one pair of adjacent Birch trees) $=$ $\#$ (configurations with one pair) $-$ $\#$ (configurations with two pairs) $+$ $\#$ (configurations with three pairs) $-$ $\#$ (configurations with four pairs). To compute the first term, note that we can treat the adjacent pair of Birch trees as one separate tree. This then gives $\frac{11!}{3! \cdot 3! \cdot 4!}$ configurations. For the second term, we have two cases. The two pairs could either happen consecutively (BBB) or separately (BB BB). They both give $\frac{10!}{2! \cdot 3! \cdot 4!}$ cases. So our second term is $\frac{2 \cdot 10!}{2! \cdot 3! \cdot 4!}$ The third term can also happen in two ways. The three pairs could be arranged like BBBB or BBB BB. Both cases together give $\frac{2 \cdot 9!}{3! \cdot 4!}$ arrangements. The final term can happen in one way (BBBBB). This gives $\frac{8!}{3! \cdot 4!}$ arrangements. Substituting these into our PIE expression, we find that there are $25760$ configurations with at least one pair of adjacent Birch trees. Therefore, there are a total of $\frac{12!}{3! \cdot 4! \cdot 5!} - 25760 = 1960$ configurations with no adjacent Birch trees. Thus, the probability of a given configuration having no two adjacent Birch trees is given by $\frac{1960}{\frac{12!}{3! \cdot 4! \cdot 5!}} = \frac{7}{99}$ Therefore, the desired result is given by $7+99 = \boxed{106}$

106

09fd4b9b449758113ff8ea1d9eeaf004

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

Here is a solution leaving out nothing. This solution is dedicated to those that are in self study and wish to learn the most they can. I will make it as elementary as possible and intuition based. Arrange first the $3$ maple and $4$ oaks as $MMMOOOO$ . We then notice that for none of the $5$ birch trees to be adjacent, they must be put in between these $M$ 's and $O$ 's. We then see that there are $8$ spots to put these $5$ birch trees in. So we can select $5$ spots for these birch trees in $\binom{8}{5}$ . But then, we can rearrange the $M$ 's and $O$ 's in $7!/(3!4!)=\binom{7}{3}$ ways. So then there are $\binom{8}{5}\binom{7}{3}$ valid arrangements with no given consecutive birch trees. There are then a total of $\frac{12!}{3!4!5!}$ different total arrangements. Therefore the probability is given as $\frac{\binom{8}{5}\binom{7}{3}}{\frac{12!}{3!4!5!}}=\frac{7}{99}$ , so the answer is $7+99=\boxed{106}$

106

44abeb1219f1dc9c4d342fc2b906cd0b

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12

function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

If $f(2+x)=f(2-x)$ , then substituting $t=2+x$ gives $f(t)=f(4-t)$ . Similarly, $f(t)=f(14-t)$ . In particular, \[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\] Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\pmod{10}$ are also roots. To see that these may be the only integer roots, observe that the function \[f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}\] satisfies the conditions and has no other roots. In the interval $-1000\leq x\leq 1000$ , there are $201$ multiples of $10$ and $200$ numbers that are congruent to $4 \pmod{10}$ , therefore the minimum number of roots is $\boxed{401}$

401

44abeb1219f1dc9c4d342fc2b906cd0b

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12

function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

We notice that the function has reflectional symmetry across both $x=2$ and $x=7$ . We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\pm 10$ as roots. Continuing this shows that the roots are $0 \mod 10$ or $4 \mod 10$ . There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of $\boxed{401}$ $QED \blacksquare$

401

44abeb1219f1dc9c4d342fc2b906cd0b

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12

function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only counting the total number of x values so ignore the double counted values. This means that $x = 0, \pm 5, \pm 10, \pm 15... \pm 1000$ so the answer is 400 + 1 = $\boxed{401}$

401

44abeb1219f1dc9c4d342fc2b906cd0b

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12

function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

Let $z$ be an arbitrary zero. If $z=2-x$ , then $x=2-z$ and $2+x=4-z$ . Repeat with other equation to find if $z$ is a zero then so are $4-z$ and $14-z$ . From $0$ , we get $4$ and $14$ . Now note that applying either of these twice will return $z$ , so we must apply them in an alternating fashion for distinct roots. Doing so to $4$ and $14$ returns $10$ and $-10$ , respectively. A pattern will emerge of each path hitting a multiple of $10$ after $2$ moves. Hence, we will reach $\pm 1000$ after $200$ jumps in either direction. Including zero, there are $2\cdot200+1=\boxed{401}$

401

d26d40c2eba5a703f1c398731324d8be

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_13

Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$

We know that $\tan(\arctan(x)) = x$ so we can repeatedly apply the addition formula, $\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ . Let $a = \cot^{-1}(3)$ $b=\cot^{-1}(7)$ $c=\cot^{-1}(13)$ , and $d=\cot^{-1}(21)$ . We have so and so Thus our answer is $10\cdot\frac{3}{2}=\boxed{015}$

015

d26d40c2eba5a703f1c398731324d8be

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_13

Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$

On the coordinate plane, let $O=(0,0)$ $A_1=(3,0)$ $A_2=(3,1)$ $B_1=(21,7)$ $B_2=(20,10)$ $C_1=(260,130)$ $C_2=(250,150)$ $D_1=(5250,3150)$ $D_2=(5100,3400)$ , and $H=(5100,0)$ . We see that $\cot^{-1}(\angle A_2OA_1)=3$ $\cot^{-1}(\angle B_2OB_1)=7$ $\cot^{-1}(\angle C_2OC_1)=13$ , and $\cot^{-1}(\angle D_2OD_1)=21$ . The sum of these four angles forms the angle of triangle $OD_2H$ , which has a cotangent of $\frac{5100}{3400}=\frac{3}{2}$ , which must mean that $\cot( \cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21)=\frac{3}{2}$ . So the answer is $10\cdot\left(\frac{3}{2}\right)=\boxed{015}.$

015

d26d40c2eba5a703f1c398731324d8be

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_13

Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$

Recall that $\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta$ and that $\arg(a + bi) = \tan^{-1}\frac{b}{a}$ . Then letting $w = 1 + 3i, x = 1 + 7i, y = 1 + 13i,$ and $z = 1 + 21i$ , we are left with \[10\cot(\frac{\pi}{2} - \arg w + \frac{\pi}{2} - \arg x + \frac{\pi}{2} - \arg y + \frac{\pi}{2} - \arg z) = 10\cot(2\pi - \arg wxyz)\] \[= -10\cot(\arg wxyz).\] Expanding $wxyz$ , we are left with \[(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)\] \[= (2+i)(13+i)(21+i)\] \[= (25+15i)(21+i)\] \[= (5+3i)(21+i)\] \[= (102+68i)\] \[= (3+2i)\] \[= 10\cot \tan^{-1}\frac{2}{3}\] \[= 10 \cdot \frac{3}{2} = \boxed{015}\]

015

0134a4f3ab44acdc7cc5f2253a49fafb

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_14

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Take an even positive integer $x$ $x$ is either $0 \bmod{6}$ $2 \bmod{6}$ , or $4 \bmod{6}$ . Notice that the numbers $9$ $15$ $21$ , ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases: If $x \ge 18$ and is $0 \bmod{6}$ $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$ . Note that $9$ and $9+6n$ are both odd composites. If $x\ge 44$ and is $2 \bmod{6}$ $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$ . Note that $35$ and $9+6n$ are both odd composites. If $x\ge 34$ and is $4 \bmod{6}$ $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$ . Note that $25$ and $9+6n$ are both odd composites. Clearly, if $x \ge 44$ , it can be expressed as a sum of 2 odd composites. However, if $x = 42$ , it can also be expressed using case 1, and if $x = 40$ , using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$ . Therefore, $\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.

038

0134a4f3ab44acdc7cc5f2253a49fafb

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_14

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$ , then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$ , which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$ , yielding a maximal answer of 38. Since $38-25=13$ , which is prime, the answer is $\boxed{038}$

038

0134a4f3ab44acdc7cc5f2253a49fafb

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_14

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

The easiest method is to notice that any odd number that ends in a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35... For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9(ex. 34, 44 . . .). Hence the largest number that ends with a 4 that satisfies the conditions is 14. If you list out all the numbers(15, 27, 9, 21, 33), you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and any number greater that ends with a 3 is bad(ex. 58, 68. . .), so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be $\boxed{38}$

38

0134a4f3ab44acdc7cc5f2253a49fafb

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_14

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Claim: The answer is $\boxed{038}$

038

0134a4f3ab44acdc7cc5f2253a49fafb

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_14

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

All numbers that could possibly work must be $2 \cdot p$ where $p$ is prime. As previous solutions stated, the maximum number that could possibly work by Chicken McNugget is $9 \cdot 25 - 9 - 25 = 225-34 = 191$ . We then bash from top to bottom: 1. $178 = 89 \cdot 2 => 87 + 91$ - refuted 2. $166 = 83 \cdot 2 => 81 + 85$ - refuted 3. $158 = 79 \cdot 2 => 77 + 81$ - refuted 4. $146 = 73 \cdot 2 => 69 + 77$ - refuted 5. $142 = 71 \cdot 2 => 65 + 77$ - refuted 6. $134 = 67 \cdot 2 => 65 + 69$ - refuted 7. $122 = 61 \cdot 2 => 57 + 65$ - refuted 8. $118 = 59 \cdot 2 => 55 + 63$ - refuted 9. $106 = 53 \cdot 2 => 51 + 55$ - refuted 10. $94 = 47 \cdot 2 => 45 + 49$ - refuted 11. $86 = 43 \cdot 2 => 35 + 51$ - refuted 12. $82 = 41 \cdot 2 => 33 + 49$ - refuted 13. $74 = 37 \cdot 2 => 35 + 39$ - refuted 14. $62 = 31 \cdot 2 => 27 + 35$ - refuted 15. $58 = 29 \cdot 2 => 25 + 33$ - refuted 16. $46 = 23 \cdot 2 => 21 + 25$ - refuted 17. $38 = 19 \cdot 2 => = 19 + 19$ - it works! Because we did a very systematic bash as shown, we are confident the answer is $\boxed{038}$

038

0134a4f3ab44acdc7cc5f2253a49fafb

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_14

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

As stated above, all numbers that could possibly work must be $2 \cdot p$ where $p$ is prime. If $p$ > 30, we consider $p$ by modulo 30. $p$ could be 1,7,11,13,17,19,23,29 modulo 30. $2 \cdot p$ can be expressed as ( $p$ $q$ )+( $p$ $q$ ) for some positive, even $q$ less then $p$ If $p$ $1 \bmod{30}$ , p±4 would both be composite If $p$ $7 \bmod{30}$ , p±2 would both be composite If $p$ $11 \bmod{30}$ , p±14 would both be composite If $p$ $13 \bmod{30}$ , p±8 would both be composite If $p$ $17 \bmod{30}$ , p±8 would both be composite If $p$ $19 \bmod{30}$ , p±14 would both be composite If $p$ $23 \bmod{30}$ , p±2 would both be composite If $p$ $29 \bmod{30}$ , p±4 would both be composite So $p$ < 30 From here, just try all possible p and find the answer is $\boxed{038}$

038

549ceac9356cc274be0ef7fbfa9461e3

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_15

Determine $x^2+y^2+z^2+w^2$ if

Rewrite the system of equations as \[\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.\] This equation is satisfied when $t \in \{4, 16, 36, 64\}$ . After clearing fractions, for each of the values $t=4,16,36,64$ , we have the equation \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\] where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$ , for $k=1,3,5,7$ Since the polynomials on each side are equal at $t=4,16,36,64$ , we can express the difference of the two polynomials by a quartic polynomial that has roots at $t=4,16,36,64$ , so \begin{align} \tag{\dag}x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)-F(t) = -(t-4)(t-16)(t-36)(t-64) \end{align} The leading coefficient of the RHS is $-1$ because the leading coefficient of the LHS is $-1$ Plug in $t=1^2, 3^2, 5^2, 7^2$ in succession, into $(\dag)$ . In each case, most terms drop, and we end up with \begin{align*} x^2=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}, \quad y^2=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}},\quad z^2=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}},\quad w^2=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}} \end{align*} Adding them up we get the sum as $3^2\cdot 4=\boxed{036}$

036

549ceac9356cc274be0ef7fbfa9461e3

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_15

Determine $x^2+y^2+z^2+w^2$ if

As in Solution 1, we have \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\] where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$ , for $k=1,3,5,7$ Now the coefficient of $t^3$ on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the $t^4$ and $t^3$ terms are, so we can eventually apply Vieta's. We can write the long equation as \[(x^2 + y^2 + z^2 + w^2)t^3 + \dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \dots\] Rearranging gives us \[t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2)t^3 \dots = 0.\] By Vieta's, we know that the sum of the roots of this equation is \[1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.\] (recall that the roots of the original and this manipulated form of it had roots $2^2, 4^2, 6^2,$ and $8^2$ ). Thus, \[x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \boxed{36}.\]

36

549ceac9356cc274be0ef7fbfa9461e3

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_15

Determine $x^2+y^2+z^2+w^2$ if

Notice how on each line, we have equations of the form $\frac{x^2}{a-1^2}+\frac{y^2}{a-3^2}+\frac{z^2}{a-5^2}+\frac{w^2}{a-7^2}=1$ . We can let this be a polynomial, with respect to $a$ . We can say that $w^2$ $x^2$ $y^2$ , and $z^2$ are coefficients, and not variables. So, we can now expand the fractions to get $(a-1)(a-9)(a-25)(a-49)=x^2(a-9)(a-25)(a-49)$ $+ y^2(a-1)(a-25)(a-49)$ $+ z^2(a-1)(a-3)(a-7)$ $+ w^2(a-1)(a-9)(a-25)$ Now, we have arrived at this huge expression, but what do we do with it? Well, we can look at what we want to find - $x^2+y^2+z^2+w^2$ . So, we want the sum of $x^2$ $y^2$ $z^2$ , and $w^2$ . Looking back to our expression, we can note how on the right hand side, the $a^3$ terms add to $x^2+y^2+z^2+w^2$ . Also, on the left hand side, the $a^3$ coefficient is $84$ (which is achievable by Vieta's formulas rather than expanding if you want to save a few seconds). So, moving all the $a^3$ terms to the left hand side, then we have that by Vieta's formulas, the sum of the roots is $-84-x^2-y^2-z^2-w^2=-(2^2+4^2+6^2+8^2)$ . Then, we can solve to find that $x^2+y^2+z^2+w^2=120-84=\boxed{036}$

036

a3d97b73187d5309b4a12258ad12e85e

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_1

Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms. $x^{24}=w$ $y^{40}=w$ , and $(xyz)^{12}=w$ . If we now convert everything to a power of $120$ , it will be easy to isolate $z$ and $w$ $x^{120}=w^5$ $y^{120}=w^3$ , and $(xyz)^{120}=w^{10}$ With some substitution, we get $w^5w^3z^{120}=w^{10}$ and $\log_zw=\boxed{060}$

060

a3d97b73187d5309b4a12258ad12e85e

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_1

Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

First we'll convert everything to exponential form. $x^{24}=w$ $y^{40}=w$ , and $(xyz)^{12}=w$ . The only expression containing $z$ is $(xyz)^{12}=w$ . It now becomes clear that one way to find $\log_z w$ is to find what $x^{12}$ and $y^{12}$ are in terms of $w$ Taking the square root of the equation $x^{24}=w$ results in $x^{12}=w^{\frac{1}{2}}$ . Raising both sides of $y^{40}=w$ to the $\frac{12}{40}$ th power gives $y^{12}=w^{\frac{3}{10}}$ Going back to $(xyz)^{12}=w$ , we can substitute the $x^{12}$ and $y^{12}$ with $w^{1/2}$ and $w^{3/10}$ , respectively. We now have $w^{1/2}w^{3/10}z^{12}=w$ . Simplifying, we get $z^{60}=w$ . So our answer is $\boxed{060}$

060

a3d97b73187d5309b4a12258ad12e85e

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_1

Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

Applying the change of base formula, \begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\ \log_y w = 40 &\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\ \log_{xyz} w = 12 &\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*} Therefore, $\frac {\log z}{\log w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$ Hence, $\log_z w = \boxed{060}$

060

a3d97b73187d5309b4a12258ad12e85e

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_1

Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

Since $\log_a b = \frac{1}{\log_b a}$ , the given conditions can be rewritten as $\log_w x = \frac{1}{24}$ $\log_w y = \frac{1}{40}$ , and $\log_w xyz = \frac{1}{12}$ . Since $\log_a \frac{b}{c} = \log_a b - \log_a c$ $\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}$ . Therefore, $\log_z w = \boxed{060}$

060

a3d97b73187d5309b4a12258ad12e85e

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_1

Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

If we convert all of the equations into exponential form, we receive $x^{24}=w$ $y^{40}=w$ , and $(xyz)^{12}=w$ . The last equation can also be written as $x^{12}y^{12}z^{12}=w$ . Also note that by multiplying the first two equations, we get, $x^{24}y^{40}= w^{2}$ . Taking the square root of this, we find that $x^{12}y^{20}=w$ . Recall, $x^{12}y^{12}z^{12}=w$ . Thus, $z^{12}= y^{8}$ . Also recall, $y^{40}=w$ . Therefore, $z^{60}$ $y^{40}$ $w$ . So, $\log_z w$ $\boxed{060}$

060

a3d97b73187d5309b4a12258ad12e85e

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_1

Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

Converting all of the logarithms to exponentials gives $x^{24} = w, y^{40} =w,$ and $x^{12}y^{12}z^{12}=w.$ Thus, we have $y^{40} = x^{24} \Rightarrow z^3=y^2.$ We are looking for $\log_z w,$ which by substitution, is $\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.$

60

835b447544c3cd794c6f4e93ac55af96

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_2

Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$

It is best to get rid of the absolute values first. Under the given circumstances, we notice that $|x-p|=x-p$ $|x-15|=15-x$ , and $|x-p-15|=15+p-x$ Adding these together, we find that the sum is equal to $30-x$ , which attains its minimum value (on the given interval $p \leq x \leq 15$ ) when $x=15$ , giving a minimum of $\boxed{015}$

015

835b447544c3cd794c6f4e93ac55af96

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_2

Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$

Let $p$ be equal to $15 - \varepsilon$ , where $\varepsilon$ is an almost neglectable value. Because of the small value $\varepsilon$ , the domain of $f(x)$ is basically the set ${15}$ . plugging in $15$ gives $\varepsilon + 0 + 15 - \varepsilon$ , or $15$ , so the answer is $\boxed{15}$

15

55d16eea30af4e724635ff729b5b76a4

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_3

What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$

If we were to expand by squaring, we would get a quartic polynomial , which isn't always the easiest thing to deal with. Instead, we substitute $y$ for $x^2+18x+30$ , so that the equation becomes $y=2\sqrt{y+15}$ Now we can square; solving for $y$ , we get $y=10$ or $y=-6$ . The second root is extraneous since $2\sqrt{y+15}$ is always non-negative (and moreover, plugging in $y=-6$ , we get $-6=6$ , which is obviously false). Hence we have $y=10$ as the only solution for $y$ . Substituting $x^2+18x+30$ back in for $y$ Both of the roots of this equation are real, since its discriminant is $18^2 - 4 \cdot 1 \cdot 20 = 244$ , which is positive. Thus by Vieta's formulas , the product of the real roots is simply $\boxed{020}$

020

55d16eea30af4e724635ff729b5b76a4

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_3

What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$

We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: \[(x^2+ 18x + 45) - 2\sqrt{x^2+18x+45} - 15 = 0.\] Letting $n = \sqrt{x^2+18x+45}$ , we have $n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0$ . Because the square root of a real number can't be negative, the only possible $n$ is $5$ Substituting that in, we have \[\sqrt{x^2+18x+45} = 5 \Longrightarrow x^2 + 18x + 45 = 25 \Longrightarrow x^2+18x+20=0.\] Reasoning as in Solution 1, the product of the roots is $\boxed{020}$

020

55d16eea30af4e724635ff729b5b76a4

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_3

What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$

Begin by completing the square on both sides of the equation, which gives \[(x+9)^2-51=2\sqrt{(x+3)(x+15)}\] Now by substituting $y=x+9$ , we get $y^2-51=2\sqrt{(y-6)(y+6)}$ , or \[y^4-106y^2+2745=0\] The solutions in $y$ are then \[y=x+9=\pm3\sqrt{5},\pm\sqrt{61}\] Turns out, $\pm3\sqrt{5}$ are a pair of extraneous solutions. Thus, our answer is then \[\left(\sqrt{61}-9\right)\left(-\sqrt{61}-9\right)=81-61=\boxed{020}\] By difference of squares.

020

55d16eea30af4e724635ff729b5b76a4

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_3

What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$

We are given the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] Squaring both sides yields \[(x^2+18x+30)^2=4(x^2+18x+45)\] \[(x^2+18x+30)^2=4(x^2+18x+30+15)\] \[(x^2+18x+30)^2=4(x^2+18x+30)+60\] \[(x^2+18x+30)^2-4(x^2+18x+30)-60=0\] Substituting $y=x^2+18x+30$ yields \[y^2-4y-60=0\] \[(y+6)(y-10)=0\] Thus $y=x^2+18x+30=-6,10$ . However if $y=-6$ , the left side of the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] would be negative while the right side is negative. Thus $y=10$ is the only possible value and we have \[x^2+18x+30=10\] \[x^2+18x+20=0\] Since the discriminant $\sqrt{18^2-4\cdot20}$ is real, both the roots are real. Thus by Vieta's Formulas, the product of the roots is the constant, $\boxed{20}$

20

6055d45bfd28aeaed4616d2168f71cfe

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_4

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

Because we are given a right angle, we look for ways to apply the Pythagorean Theorem . Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$ . Let $OE=x$ and $OD=y$ . We're trying to find $x^2+y^2$ Applying the Pythagorean Theorem, $OA^2 = OD^2 + AD^2$ and $OC^2 = EC^2 + EO^2$ Thus, $\left(\sqrt{50}\right)^2 = y^2 + (6-x)^2$ , and $\left(\sqrt{50}\right)^2 = x^2 + (y+2)^2$ . We solve this system to get $x = 1$ and $y = 5$ , such that the answer is $1^2 + 5^2 = \boxed{026}$

026

6055d45bfd28aeaed4616d2168f71cfe

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_4

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

Drop perpendiculars from $O$ to $AB$ (with foot $T_1$ ), $M$ to $OT_1$ (with foot $T_2$ ), and $M$ to $AB$ (with foot $T_3$ ). Also, mark the midpoint $M$ of $AC$ First notice that by computation, $OAC$ is a $\sqrt {50} - \sqrt {40} - \sqrt {50}$ isosceles triangle, so $AC = MO$ . Then, notice that $\angle MOT_2 = \angle T_3MO = \angle BAC$ . Therefore, the two blue triangles are congruent, from which we deduce $MT_2 = 2$ and $OT_2 = 6$ . As $T_3B = 3$ and $MT_3 = 1$ , we subtract and get $OT_1 = 5,T_1B = 1$ . Then the Pythagorean Theorem tells us that $OB^2 = \boxed{026}$

026

6055d45bfd28aeaed4616d2168f71cfe

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_4

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

Draw segment $OB$ with length $x$ , and draw radius $OQ$ such that $OQ$ bisects chord $AC$ at point $M$ . This also means that $OQ$ is perpendicular to $AC$ . By the Pythagorean Theorem, we get that $AC=\sqrt{(BC)^2+(AB)^2}=2\sqrt{10}$ , and therefore $AM=\sqrt{10}$ . Also by the Pythagorean theorem, we can find that $OM=\sqrt{50-10}=2\sqrt{10}$ Next, find $\angle BAC=\arctan{\left(\frac{2}{6}\right)}$ and $\angle OAM=\arctan{\left(\frac{2\sqrt{10}}{\sqrt{10}}\right)}$ . Since $\angle OAB=\angle OAM-\angle BAC$ , we get \[\angle OAB=\arctan{2}-\arctan{\frac{1}{3}}\] \[\tan{(\angle OAB)}=\tan{(\arctan{2}-\arctan{\frac{1}{3}})}\] By the subtraction formula for $\tan$ , we get \[\tan{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}\] \[\tan{(\angle OAB)}=1\] \[\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}\] Finally, by the Law of Cosines on $\triangle OAB$ , we get \[x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}\] \[x^2=\boxed{026}.\]

026

6055d45bfd28aeaed4616d2168f71cfe

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_4

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

We use coordinates. Let the circle have center $(0,0)$ and radius $\sqrt{50}$ ; this circle has equation $x^2 + y^2 = 50$ . Let the coordinates of $B$ be $(a,b)$ . We want to find $a^2 + b^2$ $A$ and $C$ with coordinates $(a,b+6)$ and $(a+2,b)$ , respectively, both lie on the circle. From this we obtain the system of equations $a^2 + (b+6)^2 = 50$ $(a+2)^2 + b^2 = 50$ After expanding these terms, we notice by subtracting the first and second equations, we can cancel out $a^2$ and $b^2$ . after substituting $a=3b+8$ and plugging back in, we realize that $(a,b)=(-7,-5)$ or $(5,-1)$ . Since the first point is out of the circle, we find that $(5,-1)$ is the only relevant answer. This paragraph is written by ~hastapasta. Solving, we get $a=5$ and $b=-1$ , so the distance is $a^2 + b^2 = \boxed{026}$

026

6055d45bfd28aeaed4616d2168f71cfe

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_4

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

I will use the law of cosines in triangle $\triangle OAC$ and $\triangle OBC$ $AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 2^2} = 2 \sqrt{10}$ $\cos \angle ACB = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}}$ $\cos \angle ACO = \frac{AC^2+OC^2-OA^2}{2 \cdot AC \cdot OC} = \frac{(2\sqrt{10})^2+(\sqrt{50})^2-(\sqrt{50})^2}{2 \cdot 2\sqrt{10} \cdot \sqrt{50}} = \frac{1}{\sqrt{5}}$ $\sin \angle ACB = \sqrt{1-\cos^2 \angle ACB} = \sqrt{1-(\frac{1}{\sqrt{10}})^2} = \frac{3}{\sqrt{10}}$ $\sin \angle ACO = \sqrt{1-\cos^2 \angle ACO} = \sqrt{1-(\frac{1}{\sqrt{5}})^2} = \frac{2}{\sqrt{5}}$ $\cos \angle OCB = \cos (\angle ACB - \angle ACO) = \cos \angle ACB \cdot \cos \angle ACO + \sin \angle ACB \cdot \sin \angle ACO = \frac{1}{\sqrt{10}} \cdot \frac{1}{\sqrt{5}} + \frac{3}{\sqrt{10}} \cdot \frac{2}{\sqrt{5}} = \frac{7}{5\sqrt{2}}$ $OB^2 = OC^2 + BC^2 - 2 \cdot OC \cdot BC \cdot \cos \angle OCB = (\sqrt{50})^2 + 2^2 - 2 \cdot \sqrt{50} \cdot 2 \cdot \frac{7}{5\sqrt{2}} = 50 + 4 - 28 = \boxed{026}$

026

6055d45bfd28aeaed4616d2168f71cfe

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_4

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

Notice that $50=5^2+5^2=7^2+1^2$ , and by the size of the diagram, it seems reasonable that $OA$ represents $5^2+5^2$ , and $OC$ means the $7^1+1^2$ , and indeed, the values work ( $7-5=2$ and $5+1=6$ ), so $OB^2=5^2+1^2=\boxed{026}$

026

799392f5f29c99c6764f167e2a057bd7

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_5

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

One way to solve this problem is by substitution . We have $x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10$ Hence observe that we can write $w=x+y$ and $z=xy$ This reduces the equations to $w^2-2z=7$ and $w(7-z)=10$ Because we want the largest possible $w$ , let's find an expression for $z$ in terms of $w$ $w^2-7=2z \implies z=\frac{w^2-7}{2}$ Substituting, $w^3-21w+20=0$ , which factorizes as $(w-1)(w+5)(w-4)=0$ (the Rational Root Theorem may be used here, along with synthetic division). The largest possible solution is therefore $x+y=w=\boxed{004}$

004

799392f5f29c99c6764f167e2a057bd7

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_5

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

An alternate way to solve this is to let $x=a+bi$ and $y=c+di$ Because we are looking for a value of $x+y$ that is real, we know that $d=-b$ , and thus $y=c-bi$ Expanding $x^2+y^2=7+0i$ will give two equations, since the real and imaginary parts must match up. $(a+bi)^2+(c-bi)^2=7+0i$ $(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i$ Looking at the imaginary part of that equation, $2ab-2cb=0$ , so $a=c$ , and $x$ and $y$ are actually complex conjugates. Looking at the real part of the equation and plugging in $a=c$ $2a^2-2b^2=7$ , or $2b^2=2a^2-7$ Now, evaluating the real part of $(a+bi)^3+(a-bi)^3$ , which equals $10$ (ignoring the odd powers of $i$ , since they would not result in something in the form of $10+0i$ ): $a^3+3a(bi)^2+a^3+3a(-bi)^2=10$ $2a^3-6ab^2=10$ Since we know that $2b^2=2a^2-7$ , it can be plugged in for $b^2$ in the above equation to yield: $2a^3-3a(2a^2-7)=10$ $-4a^3+21a=10$ $4a^3-21a+10=0$ Since the problem is looking for $x+y=2a$ to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, $a=10, a=5, a=\frac{5}{2}$ all fail, but $a=2$ does work. Thus, the real part of both numbers is $2$ , and their sum is $\boxed{004}$

004

799392f5f29c99c6764f167e2a057bd7

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_5

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

Begin by assuming that $x$ and $y$ are roots of some polynomial of the form $w^2+bw+c$ , such that by Vieta's Formulæ and some algebra (left as an exercise to the reader), $b^2-2c=7$ and $3bc-b^3=10$ . Substituting $c=\frac{b^2-7}{2}$ , we deduce that $b^3-21b-20=0$ , whose roots are $-4$ $-1$ , and $5$ . Since $-b$ is the sum of the roots and is maximized when $b=-4$ , the answer is $-(-4)=\boxed{004}$

004

799392f5f29c99c6764f167e2a057bd7

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_5

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

$x^3 + y^3 = 10 = (x+y)(x^2-xy+y^2) = (x+y)(7-xy) \implies xy = 7 - \frac{10}{x+y}.$ Also, $(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 = 10 + 3xy(x+y).$ Substituting our above into this, we get $10 + 3(7-\frac{10}{x+y})(x+y) = 21x+21y-20 = (x+y)^3$ . Letting $p = x+y$ , we have that $p^3 - 21p + 20 = 0$ . Testing $p = 1$ , we find that this is a root, to get $(p-1)(p^2+p-20) = 0 \implies p = -5, 1, 4 \implies \boxed{4}$

4

a8aedee36646aa980459b4330dd65c1a

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_6

Let $a_n=6^{n}+8^{n}$ . Determine the remainder upon dividing $a_ {83}$ by $49$

Firstly, we try to find a relationship between the numbers we're provided with and $49$ . We notice that $49=7^2$ , and both $6$ and $8$ are greater or less than $7$ by $1$ Thus, expressing the numbers in terms of $7$ , we get $a_{83} = (7-1)^{83}+(7+1)^{83}$ Applying the Binomial Theorem , half of our terms cancel out and we are left with $2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)$ . We realize that all of these terms are divisible by $49$ except the final term. After some quick division, our answer is $\boxed{035}$

035

a8aedee36646aa980459b4330dd65c1a

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_6

Let $a_n=6^{n}+8^{n}$ . Determine the remainder upon dividing $a_ {83}$ by $49$

As the value of $a$ is obviously $6^{83}+8^{83}$ we look for a pattern with others. With a bit of digging, we discover that $6^n+6^m$ where $m$ and $n$ are odd is equal to $\boxed{35}\:49$

35

460baf96298d79203aa02271ecd1ac8f

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

We can use complementary counting , by finding the probability that none of the three knights are sitting next to each other and subtracting it from $1$ Imagine that the $22$ other (indistinguishable) people are already seated, and fixed into place. We will place $A$ $B$ , and $C$ with and without the restriction. There are $22$ places to put $A$ , followed by $21$ places to put $B$ , and $20$ places to put $C$ after $A$ and $B$ . Hence, there are $22\cdot21\cdot20$ ways to place $A, B, C$ in between these people with restrictions. Without restrictions, there are $22$ places to put $A$ , followed by $23$ places to put $B$ , and $24$ places to put $C$ after $A$ and $B$ . Hence, there are $22\cdot23\cdot24$ ways to place $A,B,C$ in between these people without restrictions. Thus, the desired probability is $1-\frac{22\cdot21\cdot20}{22\cdot23\cdot24}=1-\frac{420}{552}=1-\frac{35}{46}=\frac{11}{46}$ , and the answer is $11+46=\boxed{057}$

057

460baf96298d79203aa02271ecd1ac8f

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

There are $(25-1)! = 24!$ configurations for the knights about the table (since configurations that are derived from each other simply by a rotation are really the same, and should not be counted multiple times). There are ${3\choose 2} = 3$ ways to pick a pair of knights from the trio, and there are $2! = 2$ ways to determine in which order they are seated. Since these two knights must be together, we let them be a single entity, so there are $(24-1)! = 23!$ configurations for the entities. However, this overcounts the instances in which the trio sits together; when all three knights sit together, then two of the pairs from the previous case are counted. However, we only want to count this as one case, so we need to subtract the number of instances in which the trio sits together (as a single entity). There are $3! = 6$ ways to determine their order, and there are $(23-1)! = 22!$ configurations. Thus, the required probability is $\frac{2 \times 3 \times 23! - 6 \times 22!}{24!} = \frac{11}{46}$ , and the answer is $\boxed{057}$

057

460baf96298d79203aa02271ecd1ac8f

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Number the knights around the table from $1$ to $25$ . The total number of ways to pick the knights is \[{25\choose 3} = \frac{25\cdot24\cdot23}{3\cdot2\cdot1} = 25\cdot23\cdot4\] There are two possibilities: either all three sit next to each other, or two sit next to each other and one is not sitting next to the other two. Case 1: All three sit next to each other. In this case, you are picking $(1,2,3)$ $(2,3,4)$ $(3,4,5)$ $(4,5,6)$ , ..., $(23,24,25)$ $(24,25,1)$ $(25,1,2)$ . This makes $25$ combinations. Case 2: Like above, there are $25$ ways to pick the pair of knights sitting next to each other. Once a pair is picked, you cannot pick either of the two adjacent knights. (For example, if you pick $(5,6)$ , you may not pick 4 or 7.) Thus, there are $25-4=21$ ways to pick the third knight, for a total of $25\cdot21$ combinations. Thus, you have a total of $25 + (25\cdot21) = 25\cdot22$ allowable ways to pick the knights. The probability is $\frac{25\cdot22}{25\cdot23\cdot4} = \frac{11}{46}$ , and the answer is $\boxed{057}$

057

460baf96298d79203aa02271ecd1ac8f

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Pick an arbitrary spot for the first knight. Then pick spots for the next two knights in order. Case 1: The second knight sits next to the first knight. There are $2$ possible places for this out of $24$ , so the probability of this is $\frac{1}{12}$ . We do not need to consider the third knight. Case 2: The second knight sits two spaces apart from the first knight. There are $2$ possible places for this out of $24$ , so the probability is $\frac{1}{12}$ . Then there are $3$ places out of a remaining $23$ for the third knight to sit, so the total probability for this case is $\frac{1}{12} \times \frac{3}{23}$ Case 3: The second knight sits three or more spaces apart from the first knight. There are $20$ possible places for this out of $24$ , so the probability is $\frac{5}{6}$ . Then there are $4$ places to put the last knight out of $23$ , so the total probability for this case is $\frac{5}{6}\times\frac{4}{23}$ Now we add the probabilities to get the total: \[\frac{1}{12}+\frac{1}{12}\times\frac{3}{23}+\frac{5}{6}\times\frac{4}{23}=\frac{1}{12}\times\frac{1}{23}\left(23+3+40\right)=\frac{66}{12\times 23}\] \[=\frac{6\times 11}{6\times 2 \times 23}=\frac{11}{46}\] so the answer is $\boxed{057}$

057

460baf96298d79203aa02271ecd1ac8f

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

We simplify this problem by using complementary counting and fixing one knight in place. Then, either a knight can sit two spaces apart from the fixed knight, or a knight can sit more than two spaces apart from the fixed knight. The probability is then $\frac{24\left(23\right)-\left[2\left(20\right)+20\left(19\right)\right]}{24\left(23\right)}=\frac{11}{46}$ , so the answer is $11+46=\boxed{057}$

057

460baf96298d79203aa02271ecd1ac8f

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Let $K_1, K_2, K_3$ be the knights in clockwise order. Let $A$ be the distance between $K_1$ and $K_2$ $B$ the distance between $K_2$ and $K_3$ and $C$ the distance between $K_3$ and $K_1$ $A + B + C = 25$ and $A, B, C \geq 1$ . In order to use stars and bars the numbers must be greater than or equal to 0 instead of 1, so we define $A_1 = A - 1, B_1 = B - 1, C_1 = C - 1$ $A_1 + B_1 + C_1 = 22$ , so by stars and bars there are $\binom{22 + 3 - 1}{3 - 1} = 276$ possibilities. The condition is not satisfied if $A, B, C \geq 2$ , so we can use complementary counting. Let $A_2 = A - 2, B_2 = B - 2, C_2 = C - 2$ $A_2 + B_2 + C_2 = 19$ , and by stars and bars there are $\binom{19 + 3 - 1}{3 - 1} = 210$ possibilities. This means there are $276 - 210 = 66$ possibilities where the condition is satisfied, so the probability is $\frac{11}{46}$ , resulting in $\boxed{057}$

057

460baf96298d79203aa02271ecd1ac8f

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

There are $\binom{25}{3}=\frac{25 \cdot 24 \cdot 23}{3 \cdot 2}=2300$ ways to chose $3$ knights out of $25$ knights. To ensure at least $2$ adjacent knights are chosen, first choose $1$ of the $25$ pairs of adjacent knights. After choosing the adjacent pair of knights, there are $23$ ways to choose the third knight. There are $25$ choices of $3$ knights sitting consecutively, which are counted twice. For example, choosing $(1,2)$ first, then choosing $3$ is the same as choosing $(2,3)$ first, then choosing $1$ . Therefore, there are $25 \cdot 23 -25=550$ ways to choose $3$ knights where at least $2$ of the $3$ had been sitting next to each other. \[P=\frac{550}{2300}=\frac{11}{46}\] The answer is $11+46=\boxed{057}$

057

460baf96298d79203aa02271ecd1ac8f

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

(I need to get this out) Denominator: $\binom{25}{3}$ ways to choose three knights Numerator: $\implies$ Case 1: Two knights are next to each other, third is lonely :( $\implies$ Case 2: All three knights are next to each other Our answer is $\frac{21 \cdot 25+25}{\binom{25}{3}}$ Simplifies to: $\frac{22 \cdot 25 \cdot 6}{25 \cdot 24 \cdot 3}$ $\implies$ $\frac{11}{46}$ , and our answer is $11+46=\boxed{057}$

057

3cfe80bd6a43dd0d2a296d5b1953dc01

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_8

What is the largest $2$ -digit prime factor of the integer $n = {200\choose 100}$

Expanding the binomial coefficient , we get ${200 \choose 100}=\frac{200!}{100!100!}$ . Let the required prime be $p$ ; then $10 \le p < 100$ . If $p > 50$ , then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$ . The largest such prime is $\boxed{061}$ , which is our answer.

061

3cfe80bd6a43dd0d2a296d5b1953dc01

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_8

What is the largest $2$ -digit prime factor of the integer $n = {200\choose 100}$

We know that \[{200\choose100}=\frac{200!}{100!100!}\] Since $p<100$ , there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\frac{200!}{100!100!}$ . (Note that here we assume the minimum because as $p$ goes larger in value, the number of factors of $p$ in a number decreases,) So basically, $p$ is the largest prime number such that \[\left \lfloor\frac{200}{p}\right \rfloor>3\] Since $p<\frac{200}{3}=66.66...$ , the largest prime value for $p$ is $p=\boxed{61}$

61

9bfe5532c19dec50ff1da36115a9dc5b

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_9

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$

We can rewrite the numerator to be a perfect square by adding $-\dfrac{12x \sin x}{x \sin x}$ . Thus, we must also add back $12$ This results in $\dfrac{(3x \sin x-2)^2}{x \sin x}+12$ Thus, if $3x \sin x-2=0$ , then the minimum is obviously $12$ . We show this possible with the same methods in Solution 1; thus the answer is $\boxed{012}$

012

9bfe5532c19dec50ff1da36115a9dc5b

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_9

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$

Let $y = x\sin{x}$ and rewrite the expression as $f(y) = 9y + \frac{4}{y}$ , similar to the previous solution. To minimize $f(y)$ , take the derivative of $f(y)$ and set it equal to zero. The derivative of $f(y)$ , using the Power Rule, is $f'(y)$ $9 - 4y^{-2}$ $f'(y)$ is zero only when $y = \frac{2}{3}$ or $y = -\frac{2}{3}$ . It can further be verified that $\frac{2}{3}$ and $-\frac{2}{3}$ are relative minima by finding the derivatives at other points near the critical points, or by checking that the second derivative $f''(y)=8y^{-3}$ is positive. However, since $x \sin{x}$ is always positive in the given domain, $y = \frac{2}{3}$ . Therefore, $x\sin{x}$ $\frac{2}{3}$ , and the answer is $\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}$

012

9bfe5532c19dec50ff1da36115a9dc5b

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_9

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$

As above, let $y = x\sin{x}$ . Add $\frac{12y}{y}$ to the expression and subtract $12$ , giving $f(x) = \frac{(3y+2)^2}{y} - 12$ . Taking the derivative of $f(x)$ using the Chain Rule and Quotient Rule , we have $\frac{\text{d}f(x)}{\text{d}x} = \frac{6y(3y+2)-(3y+2)^2}{y^2}$ . We find the minimum value by setting this to $0$ . Simplifying, we have $6y(3y+2) = (3y+2)^2$ and $y = \pm{\frac{2}{3}} = x\sin{x}$ . Since both $x$ and $\sin{x}$ are positive on the given interval, we can ignore the negative root. Plugging $y = \frac{2}{3}$ into our expression for $f(x)$ , we have $\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\left(\frac{2}{3}\right)}-12 = \boxed{012}$

012

9bfe5532c19dec50ff1da36115a9dc5b

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_9

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$

Set $\frac{9x^2\sin^2 x + 4}{x\sin x}$ equal to $y$ . Then multiply by $x\sin x$ on both sides to get $9x^2\sin^2 x + 4 = y\cdot x\sin x$ . We then subtract $yx\sin x$ from both sides to get $9x^2\sin^2 x + 4 - yx\sin x = 0$ . This looks like a quadratic so set $z= x\sin x$ and use quadratic equation on $9z^2 - yz + 4 = 0$ to see that $z = \frac{y\pm\sqrt{y^2-144}}{18}$ . We know that $y$ must be an integer and as small as it can be, so $y$ = 12. We plug this back in to see that $x\sin x = \frac{2}{3}$ which we can prove works using methods from solution 1. This makes the answer $\boxed{012}$

012

1de08e602382f9a64042ec19d6041a2a

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10

The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

Suppose that the two identical digits are both $1$ . Since the thousands digit must be $1$ , only one of the other three digits can be $1$ . This means the possible forms for the number are Because the number must have exactly two identical digits, $x\neq y$ $x\neq1$ , and $y\neq1$ . Hence, there are $3\cdot9\cdot8=216$ numbers of this form. Now suppose that the two identical digits are not $1$ . Reasoning similarly to before, we have the following possibilities: Again, $x\neq y$ $x\neq 1$ , and $y\neq 1$ . There are $3\cdot9\cdot8=216$ numbers of this form. Thus the answer is $216+216=\boxed{432}$

432

1de08e602382f9a64042ec19d6041a2a

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10

The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

Consider a sequence of $4$ digits instead of a $4$ -digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is $\frac{1}{10}$ . This means we can find all possible sequences with one digit repeated twice, and then divide by $10$ If we let the three distinct digits of the sequence be $a, b,$ and $c$ , with $a$ repeated twice, we can make a table with all possible sequences: \[\begin{tabular}{ccc} aabc & abac & abca \\ baac & baca & \\ bcaa && \\ \end{tabular}\] There are $6$ possible sequences. Next, we can see how many ways we can pick $a$ $b$ , and $c$ . This is $10(9)(8) = 720$ , because there are $10$ digits, from which we need to choose $3$ with regard to order. This means there are $720(6) = 4320$ sequences of length $4$ with one digit repeated. We divide by 10 to get $\boxed{432}$ as our answer.

432

1de08e602382f9a64042ec19d6041a2a

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10

The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

We'll use complementary counting. We will split up into $3$ cases: (1) no number is repeated, (2) $2$ numbers are repeated, and $2$ other numbers are repeated, (3) $3$ numbers are repeated, or (4) $4$ numbers are repeated. Case 1: There are $9$ choices for the hundreds digit (it cannot be $1$ ), $8$ choices for the tens digit (it cannot be $1$ or what was chosen for the hundreds digit), and $7$ for the units digit. This is a total of $9\cdot8\cdot7=504$ numbers. Case 2: One of the three numbers must be $1$ , and the other two numbers must be the same number, but cannot be $1$ (That will be dealt with in case 4). There are $3$ choices to put the $1$ , and there are $9$ choices (any number except for $1$ ) to pick the other number that is repeated, so a total of $3\cdot9=27$ numbers. Case 3: We will split it into $2$ subcases: one where $1$ is repeated $3$ times, and one where another number is repeated $3$ times. When $1$ is repeated $3$ times, then one of the digits is not $1$ . There are $9$ choices for that number, and $3$ choices for its location,so a total of $9\cdot3=27$ numbers. When a number other than $1$ is repeated $3$ times, then there are $9$ choices for the number, and you don't have any choices on where to put that number. So in Case 3 there are $27+9=36$ numbers Case 4: There is only $1$ number: $1111$ There are a total of $1000$ $4$ -digit numbers that begin with $1$ (from $1000$ to $1999$ ), so by complementary counting you get $1000-(504+27+36+1)=\boxed{432}$ numbers.

432

1de08e602382f9a64042ec19d6041a2a

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10

The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

Let us proceed by casework. Case 1: We will count the amount of numbers that have two identical digits that are not one. The thousands digit is fixed, and we are choosing two spots to hold two identical digits that are chosen from $0, 2-9$ , which is $9$ options. For the last digit, their are $8$ possibilities since it can be neither $1$ or the other number that is chosen. The final outcome is ${3}\choose{2}$ $* 9 * 8 = 216$ possibilities for this case. Case 2: The last case will be the amount of numbers that have two identical digits thare are $1$ . There are ${3}\choose{1}$ places to pick the $1$ . For the other $2$ digits, there are $9$ and $8$ options respectively. Thus, we have $3 * 9 * 8 = 216$ . Summing the two cases, we get $216 + 216 = \boxed{432}$

432

84ec0bf1dc88123be4f97077b97246f6

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11

The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid?

First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$ . The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$ , and one of the legs is $3\sqrt{2}$ . We apply the Pythagorean Theorem to deduce that the height is $6$ Next, we complete t he figure into a triangular prism, and find its volume, which is $\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432$ Now, we subtract off the two extra pyramids that we included, whose combined volume is $2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144$ Thus, our answer is $432-144=\boxed{288}$

288

84ec0bf1dc88123be4f97077b97246f6

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11

The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid?

Extend $EA$ and $FB$ to meet at $G$ , and $ED$ and $FC$ to meet at $H$ . Now, we have a regular tetrahedron $EFGH$ , which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\sqrt{2}$ . Using the formula for the volume of a regular tetrahedron, which is $V = \frac{\sqrt{2}S^3}{12}$ , where S is the side length of the tetrahedron, the volume of our original solid is: $V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}$

288

84ec0bf1dc88123be4f97077b97246f6

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11

The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid?

We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$ ; thus, we will integrate with respect to height from $0$ to $6$ , noting that each cross section of height $dh$ is a rectangle. The volume is then $\int_0^h(wl) \ \text{d}h$ , where $w$ is the width of the rectangle and $l$ is the length. We can express $w$ in terms of $h$ as $w=6\sqrt{2}-\sqrt{2}h$ since it decreases linearly with respect to $h$ , and $l=6\sqrt{2}+\sqrt{2}h$ since it similarly increases linearly with respect to $h$ . Now we solve: \[\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)\ \text{d}h =\int_0^6(72-2h^2)\ \text{d}h=72(6)-2\left(\frac{1}{3}\right)\left(6^3\right)=\boxed{288}\]

288

84ec0bf1dc88123be4f97077b97246f6

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11

The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid?

Draw an altitude from a vertex of the square base to the top edge. By using $30,60, 90$ triangle ratios, we obtain that the altitude has a length of $3 \sqrt{6}$ , and that little portion that hangs out has a length of $3\sqrt2$ . This is a triangular pyramid with a base of $3\sqrt6, 3\sqrt6, 3\sqrt2$ , and a height of $3\sqrt{2}$ . Since there are two of these, we can compute the sum of the volumes of these two to be $72$ . Now we are left with a triangular prism with a base of dimensions $3\sqrt6, 3\sqrt6, 3\sqrt2$ and a height of $6\sqrt2$ . We can compute the volume of this to be 216, and thus our answer is $\boxed{288}$

288

3f1557bfb02dd983288cd405dfb9cce7

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13

For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+1=5$ and for $\{5\}$ it is simply $5$ . Find the sum of all such alternating sums for $n=7$

Let $S$ be a non- empty subset of $\{1,2,3,4,5,6\}$ Then the alternating sum of $S$ , plus the alternating sum of $S \cup \{7\}$ , is $7$ . This is because, since $7$ is the largest element, when we take an alternating sum, each number in $S$ ends up with the opposite sign of each corresponding element of $S\cup \{7\}$ Because there are $2^{6}=64$ of these pairs of sets, the sum of all possible subsets of our given set is $64 \cdot 7$ , giving an answer of $\boxed{448}$

448