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48a22df7344d0edab73b0c1c78d26849 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_8 | If the radius of a circle is increased $100\%$ , the area is increased:
$\textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these}$ | Increasing by $100\%$ is the same as doubling the radius. If we let $r$ be the radius of the old circle, then the radius of the new circle is $2r.$
Since the area of the circle is given by the formula $\pi r^2,$ the area of the new circle is $\pi (2r)^2 = 4\pi r^2.$ The area is quadrupled, or increased by $\boxed{300}$ | C | 300 |
bab24288566c327e0ab1154aa5d9f923 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_9 | The area of the largest triangle that can be inscribed in a semi-circle whose radius is $r$ is:
$\textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2}$ | The area of a triangle is $\frac12 bh.$ To maximize the base, let it be equal to the diameter of the semi circle, which is equal to $2r.$ To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to $r.$ Therefore the area is $\frac12 \cdot 2r \cdot r = \boxed{2}$ | A | 2 |
81be900c6c9c29a8647ad9b2fbd94a91 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_16 | The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Use properties of exponents to move the squares outside the brackets use difference of squares.
\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]
Using the binomial theorem, we can see that the number of terms is $\boxed{5}$ | B | 5 |
97e603f0b9a637b72c5b661d2693bbcd | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_18 | Of the following
(1) $a(x-y)=ax-ay$
(2) $a^{x-y}=a^x-a^y$
(3) $\log (x-y)=\log x-\log y$
(4) $\frac{\log x}{\log y}=\log{x}-\log{y}$
(5) $a(xy)=ax \cdot ay$
$\textbf{(A)}\text{Only 1 and 4 are true}\qquad\\\textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\\textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\\textbf{(D)}\ \text{Only 1 and 2 are true}\qquad\\\textbf{(E)}\ \text{Only 1 is true}$ | The distributive property doesn't apply to logarithms or in the ways illustrated, and only applies to addition and subtraction. Also, $a^{x-y} = \frac{a^x}{a^y}$ , so $\boxed{1}$ | E | 1 |
1593d8af2008d5782cf0f6c75a39905e | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_20 | When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$ | Using synthetic division, we get that the remainder is $\boxed{2}$ | D | 2 |
1593d8af2008d5782cf0f6c75a39905e | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_20 | When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$ | By the remainder theorem, the remainder is equal to the expression $x^{13}+1$ when $x=1.$ This gives the answer of $\boxed{2.}$ | D | 2. |
1593d8af2008d5782cf0f6c75a39905e | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_20 | When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$ | Note that $x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)$ , so $x^{13} - 1$ is divisible by $x-1$ , meaning $(x^{13} - 1) + 2$ leaves a remainder of $\boxed{2.}$ | D | 2. |
cc5521e4762c2a50f82e6ef088e90849 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_21 | The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^{2}$ $8\text{ in}^{2}$ , and $6\text{ in}^{2}$ respectively is:
$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these}$ | If the sidelengths of the cubes are expressed as $a, b,$ and $c,$ then we can write three equations:
\[ab=12, bc=8, ac=6.\]
The volume is $abc.$ Notice symmetry in the equations. We can find $abc$ my multiplying all the equations and taking the positive square root.
\begin{align*} (ab)(bc)(ac) &= (12)(8)(6)\\ a^2b^2c^2 &= 576\\ abc &= \boxed{24} | B | 24 |
5f22b1079465e17c89abf6fd894c88f6 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_22 | Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these}$ | Without loss of generality, assume something costs $100$ dollars. Then with each successive discount, it would cost $90$ dollars, then $72$ dollars. This amounts to a total of $28$ dollars off, so the single discount would be $\boxed{28}$ | D | 28 |
15cba6f5de754971777da9e876c0222b | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_23 | A man buys a house for $10,000 and rents it. He puts $12\frac{1}{2}\%$ of each month's rent aside for repairs and upkeep; pays $325 a year taxes and realizes $5\frac{1}{2}\%$ on his investment. The monthly rent (in dollars) is:
$\textbf{(A)} \ \ 64.82\qquad\textbf{(B)} \ \ 83.33\qquad\textbf{(C)} \ \ 72.08\qquad\textbf{(D)} \ \ 45.83\qquad\textbf{(E)} \ \ 177.08$ | $12\frac{1}{2}\%$ is the same as $\frac{1}{8}$ , so the man sets one eighth of each month's rent aside, so he only gains $\frac{7}{8}$ of his rent. He also pays $325 each year, and he realizes $5.5\%$ , or $550, on his investment. Therefore he must have collected a total of $325 +$550 = $875 in rent. This was for the whole year, so he collected $\frac{875}{12}$ dollars each month as rent. This is only $\frac{7}{8}$ of the monthly rent, so the monthly rent in dollars is $\frac{875}{12}\cdot \frac{8}{7}=\boxed{83.33}$ | B | 83.33 |
78a66efd519363e3ceadc13a0fd88533 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_24 | The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$ | $x + \sqrt{x-2} = 4$ Original Equation
$\sqrt{x-2} = 4 - x$ Subtract x from both sides
$x-2 = 16 - 8x + x^2$ Square both sides
$x^2 - 9x + 18 = 0$ Get all terms on one side
$(x-6)(x-3) = 0$ Factor
$x = \{6, 3\}$
If you put down A as your answer, it's wrong. You need to check for extraneous roots.
$6 + \sqrt{6 - 2} = 6 + \sqrt{4} = 6 + 2 = 8 \ne 4$
$3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark$
There is $\boxed{1}$ | E | 1 |
78a66efd519363e3ceadc13a0fd88533 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_24 | The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$ | We can create symmetry in the equation: \[x+\sqrt{x-2} = 4\] \[x-2+\sqrt{x-2} = 2.\] Let $y = \sqrt{x-2}$ , then we have \[y^2+y-2 = 0\] \[(y+2)(y-1) = 0\] The two roots are $\sqrt{x-2} = -2, 1$
Notice, that the first root is extraneous as the range for the square root function is always the non-negative numbers (remember, negative numbers in square roots give imaginary numbers - imaginary numbers in square roots don't give negative numbers); thus, the only real root for $x$ occurs for the second root; squaring both sides and solving for $x$ gives $x=3 \Rightarrow \boxed{1}$ | E | 1 |
b783b3a52ff9a4137359a82df06e9845 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_25 | The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$ | $\log_{5}\frac{(125)(625)}{25}$ can be simplified to $\log_{5}\ (125)(25)$ since $25^2 = 625$ $125 = 5^3$ and $5^2 = 25$ so $\log_{5}\ 5^5$ would be the simplest form. In $\log_{5}\ 5^5$ $5^x = 5^5$ . Therefore, $x = 5$ and the answer is $\boxed{5}$ | D | 5 |
b783b3a52ff9a4137359a82df06e9845 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_25 | The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$ | $\log_{5}\frac{(125)(625)}{25}$ can be also represented as $\log_{5}\frac{(5^3)(5^4)}{5^2}= \log_{5}\frac{(5^7)}{5^2}= \log_{5} 5^5$ which can be solved to get $\boxed{5}$ | D | 5 |
367ee23bcd67eaf2f0f2db1b666d64b5 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_26 | If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$ | We have $b=\log_{10}{10^b}$ . Substituting, we find $\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}$ . Using $\log{a}-\log{b}=\log{\dfrac{a}{b}}$ , the left side becomes $\log_{10}{\dfrac{10^b}{n}}$ . Because $\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}$ $m=\boxed{10}$ | E | 10 |
367ee23bcd67eaf2f0f2db1b666d64b5 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_26 | If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$ | adding $\log_{10} n$ to both sides: \[\log_{10} m + \log_{10} n=b\] using the logarithm property: $\log_a {b} + \log_a {c}=\log_a{bc}$ \[\log_{10} {mn}=b\] rewriting in exponential notation: \[10^b=mn\] \[m=\boxed{10}\] ~Vndom | E | 10 |
7e95454dc1fc51d206821c757babb9c7 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_27 | A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:
$\textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37\text{ mph}$ | The car takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{30 \text{ miles }}=4 \text{ hr}$ to get from $A$ to $B$ . Also, it takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{40 \text{ miles }}=3 \text{ hr}$ to get from $B$ to $A$ . Therefore, the average speed is $\dfrac{240\text{ miles }}{7 \text{ hr}}=34\dfrac{2}{7}\text{ mph}$ , which is closest to $\boxed{34}$ | B | 34 |
6b8497181c28b13d625df7adeff4a048 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_28 | Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$ $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:
$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph} \qquad \textbf{(C)}\ 12\text{ mph} \qquad \textbf{(D)}\ 16\text{ mph} \qquad \textbf{(E)}\ 20\text{ mph}$ | Let the speed of boy $A$ be $a$ , and the speed of boy $B$ be $b$ . Notice that $A$ travels $4$ miles per hour slower than boy $B$ , so we can replace $b$ with $a+4$
Now let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=rt$ to make an equation, where we set the time to be equal: \[\frac{48}{a}=\frac{72}{a+4}\] Cross-multiplying gives $48a+192=72a$ . Isolating the variable $a$ , we get the equation $24a=192$ , so $a=\boxed{8}$ | B | 8 |
e33582148f9921d5e7bce3838ab5ab9e | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_30 | From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \text{None of these}$ | Let us represent the number of boys $b$ , and the number of girls $g$
From the first sentence, we get that $2(g-15)=b$
From the second sentence, we get $5(b-45)=g-15$
Expanding both equations and simplifying, we get $2g-30 = b$ and $5b = g+210$
Substituting $b$ for $2g-30$ , we get $5(2g-30)=g+210$ . Solving for $g$ , we get $g = \boxed{40}$ | A | 40 |
bcd47560365e468133a540eb0e79784b | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_32 | $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textbf{(D)}\ 8\text{ ft} \qquad \textbf{(E)}\ 4\text{ ft}$ | By the Pythagorean triple $(7,24,25)$ , the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \boxed{8}$ | D | 8 |
bcd47560365e468133a540eb0e79784b | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_32 | $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textbf{(D)}\ 8\text{ ft} \qquad \textbf{(E)}\ 4\text{ ft}$ | We can observe that the above setup forms a right angled triangles whose base is 7ft and whose hypotenuse is 25ft taking the height to be x ft.
\[x^2 + 7^2 = 25^2\] \[x^2 = 625 - 49\] \[x^2 = 576\] \[x = 24\]
Since the top of the ladder slipped by 4 ft the new height is $24 - 4 = 20 ft$ . The base of the ladder has moved so the new base is say $(7+y)$ . The hypotenuse remains the same at 25ft. So,
\[20^2 + (7+y)^2 = 25^2\] \[400 + 49 + y^2 + 14y = 625\] \[y^2 + 14y - 176 = 0\] \[y^2 + 22y - 8y - 176\] \[x(y+22) - 8(y+22)\] \[(y-8)(y+22)\]
Disregarding the negative solution to equation the solution to the problem is $\boxed{8}$ | D | 8 |
2403f18fe6adb2ad7dd64fcbe0f7c1d6 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_33 | The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:
$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$ | It must be assumed that the pipes have an equal height.
We can represent the amount of water carried per unit time by cross sectional area.
Cross sectional of Pipe with diameter $6 in$ \[\pi r^2 = \pi \cdot 3^2 = 9\pi\]
Cross sectional area of pipe with diameter $1 in$
\[\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}\]
So number of 1 in pipes required is the number obtained by dividing their cross sectional areas
\[\frac{9\pi}{\frac{\pi}{4}} = 36\]
So the answer is $\boxed{36}$ | D | 36 |
4eafea8883767b672c63c70b8a6fd8b0 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_35 | In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$ | The inradius is equal to the area divided by semiperimeter. The area is $\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$ . Therefore the inradius is $\boxed{4}$ | B | 4 |
4eafea8883767b672c63c70b8a6fd8b0 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_35 | In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$ | Since this is a right triangle, we have \[\frac{a+b-c}{2}=\boxed{4}\] | null | 4 |
72bae22d13b2c82284fa2e47ce9e654e | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_36 | A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?
$\textbf{(A)}\ 125\% \qquad \textbf{(B)}\ 100\% \qquad \textbf{(C)}\ 120\% \qquad \textbf{(D)}\ 80\% \qquad \textbf{(E)}\ 75\%$ | Without loss of generality, we can set the list price equal to $100$ . The merchant buys the goods for $100*.75=75$ . Let $x$ be the marked price.
We then use the equation $0.8x-75=25$ to solve for $x$ and get a marked price of $\boxed{125}$ | null | 125 |
6dbba58c38cdfe2363eea836904d69a1 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_38 | If the expression $\begin{pmatrix}a & c\\ d & b\end{pmatrix}$ has the value $ab-cd$ for all values of $a, b, c$ and $d$ , then the equation $\begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3$
$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \textbf{(C)}\ \text{Is satisified for no values of }x\qquad\\ \textbf{(D)}\ \text{Is satisfied for an infinite number of values of }x\qquad\\ \textbf{(E)}\ \text{None of these.}$ | By $\begin{pmatrix}a & c\\ d & b\end{pmatrix}=ab-cd$ , we have $2x^2-x=3$ . Subtracting $3$ from both sides, giving $2x^2-x-3=0$ . This factors to $(2x-3)(x+1)=0$ . Thus, $x=\dfrac{3}{2},-1$ , so the equation is $\boxed{2}$ | B | 2 |
baa96ad69c523059b0a7eb9e75654ed2 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_40 | The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$ | Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$ , using the L'Hôpital's rule, we have $\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2$ .
Thus, the answer is $\boxed{2}$ | D | 2 |
baa96ad69c523059b0a7eb9e75654ed2 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_40 | The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$ | The numerator of $\frac {x^2-1}{x-1}$ can be factored as $(x+1)(x-1)$ . The $x-1$ terms in the numerator and denominator cancel, so the expression is equal to $x+1$ so long as $x$ does not equal $1$ . Looking at the function's behavior near 1, we see that as $x$ approaches one, the expression approaches $\boxed{2}$ | D | 2 |
a3ca15d83e3806c6c621114b3714a4e5 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45 | The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$ | Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be $\binom{100}{2}=4950$ . However this also counts the 100 sides of the polygon, so the actual answer is $4950-100=\boxed{4850}$ | A | 4850 |
a3ca15d83e3806c6c621114b3714a4e5 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45 | The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$ | The formula for the number of diagonals of a polygon with $n$ sides is $n(n-3)/2$ . Taking $n=100$ , we see that the number of diagonals that may be drawn in this polygon is $100(97)/2$ or $\boxed{4850}$ | A | 4850 |
a2d7d27379eee5911205046c472a46e9 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_46 | In triangle $ABC$ $AB=12$ $AC=7$ , and $BC=10$ . If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:
$\textbf{(A)}\ \text{The area is doubled} \qquad\\ \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ \textbf{(D)}\ \text{The median is unchanged} \qquad\\ \textbf{(E)}\ \text{The area of the triangle is 0}$ | If you double sides $AB$ and $AC$ , they become $24$ and $14$ respectively. If $BC$ remains $10$ , then this triangle has area $0$ because ${14} + {10} = {24}$ , so two sides overlap the third side. Therefore the answer is $\boxed{0}$ | E | 0 |
90f8e608d5698ebf11846079d189ecfe | https://artofproblemsolving.com/wiki/index.php/2020_USOMO_Problems/Problem_4 | Suppose that $(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \le i < j \le 100$ and $|a_ib_j - a_j b_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs. | Let's start off with just $(a_1, b_1), (a_2, b_2)$ and suppose that it satisfies the given condition. We could use $(1, 1), (1, 2)$ for example. We should maximize the number of conditions that the third pair satisfies. We find out that the third pair should equal $(a_1+a_2, b_1+b_2)$
We know this must be true: \[|a_1b_2-a_2b_1| = 1\]
So \[a_1b_2-a_2b_1 = 1\]
We require the maximum conditions for $(a_3, b_3)$ \[|a_3b_2-a_2b_3| = 1\] \[|a_3b_1-a_1b_3| = 1\]
Then one case can be: \[a_3b_2-a_2b_3 = 1\] \[a_3b_1-a_1b_3 = -1\]
We try to do some stuff such as solving for $a_3$ with manipulations: \[a_3b_2a_1-a_2b_3a_1 = a_1\] \[a_3b_1a_2-a_1b_3a_2 = -a_2\] \[a_3(a_1b_2-a_2b_1) = a_1+a_2\] \[a_3 = a_1+a_2\] \[a_3b_2b_1-a_2b_3b_1 = b_1\] \[a_3b_1b_2-a_1b_3b_2 = -b_2\] \[b_3(a_1b_2-a_2b_1) = b_1+b_2\] \[b_3 = b_1+b_2\]
We showed that 3 pairs are a complete graph; however, 4 pairs are not a complete graph. We will now show that: \[a_4 = a_1+2a_2\] \[b_4 = b_1+2b_2\] \[|a_1b_1+2a_2b_1-a_1b_1-2a_1b_2| = 1\] \[2|a_2b_1-a_1b_2| = 1\]
This is clearly impossible because $1$ is not even and also $|a_2b_1-a_1b_2| = 1$ .
The answer is as follows: \[0+1+2+\ldots+2\] $a_1$ has $0$ subtractions that follow condition while $a_2$ has $1$ and then the rest has $2$ .
There are $n$ terms, so our answer be $2n-3$ and in case of $n=100$ that means \[\boxed{197}.\] ~Lopkiloinm | null | 197 |
90f8e608d5698ebf11846079d189ecfe | https://artofproblemsolving.com/wiki/index.php/2020_USOMO_Problems/Problem_4 | Suppose that $(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \le i < j \le 100$ and $|a_ib_j - a_j b_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs. | We claim the answer is $197$
Study the points $(0, 0), (a_i, b_i), (a_j, b_j)$ . If we let these be the vertices of a triangle, applying shoelace theorem gives us an area of $\frac{1}{2}|0\times{b_i}+{a_i}\times{b_j}+{b_i}\times{0}-0\times{a_i}-{b_i}\times{a_j}-{b_j}\times{0} = \frac{1}{2}|a_ib_j - a_j b_i| = \frac{1}{2}$ . Therefore, the triangle formed by the points $(0, 0), (a_i, b_i), (a_j, b_j)$ must have an area of $\frac{1}{2}$
Two cases follow.
Case 1: Both $(a_i, b_i), (a_j, b_j)$ have exactly one coordinate equal to $0$ .
Here, one point must be on the $x$ axis and the other on the $y$ axis in order for the triangle to have a positive area. For the area of the triangle to be $\frac{1}{2}$ , it follows that the points must be $(1, 0), (0, 1)$ in some order.
Case 2: At least one of $(a_i, b_i), (a_j, b_j)$ does not have exactly one coordinate equal to $0$ . Define $S[l]$ to be a list of lines such that each line in the list has some two lattice points that, with $(0, 0)$ , form a triangle with area $\frac{1}{2}$ . Note that for any such line that passes through such two lattice points, we may trivially generate infinite lattice points on the line that have nonnegative coordinates.
Note that lines $y=1$ and $x=1$ are included in $S[l]$ , because the points $(1, 1), (2, 1)$ serve as examples for $y=1$ and $(1, 1), (1, 2)$ serve as examples for $x=1$ . For the optimal construction, include the points $(1, 0)$ and all the points $(0, 1), (0, 2), (0, 3), ... , (0, 99)$ , in that order. In this case, every adjacent pair of points would count ( $98$ ), as well as picking $(0, 1)$ and a nonadjacent point ( $99$ ), so this would be $98+99=197$
To prove that this is the maximum, consider the case where some $n$ number of points were neither on $x=1$ nor on $y=1$ . In this case, we would be removing $n$ adjacent pairs and $n$ options to choose from after choosing $(0, 0)$ , resulting in a net loss of $2n$ . By having $n$ points on some other combination of lines in $S[l]$ , we would trivially have a maximum gain of $n-1$ pairs of points on the lines such that there are no lattice points between those pairs. Because these points are not on $x=1$ or $y=1$ , the altitude from a given point to the line formed by $(0, 0)$ and $(0, 1)$ and $(1, 0)$ is not $1$ , and so the area of the triangle cannot be $\frac{1}{2}$ . Thus, by not having all points on lines $x=1$ and $y=1$ , we cannot exceed the maximum of $197$ . Thus, $\boxed{197}$ is our answer. | null | 197 |
4eddc86a2b76bcb0824003bb66e85760 | https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_1 | Let $a,b,c,d$ be real numbers such that $b-d \ge 5$ and all zeros $x_1, x_2, x_3,$ and $x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real. Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take. | Using the hint we turn the equation into $\prod_{k=1} ^4 (x_k-i)(x_k+i) \implies P(i)P(-i) \implies (b-d-1)^2 + (a-c)^2 \implies \boxed{16}$ . This minimum is achieved when all the $x_i$ are equal to $1$ | null | 16 |
f25c446de036ffaadaf5676a0b0756b1 | https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_4 | problem_id
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
Name: Text, dtype: object | We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\boxed{6}$ | null | 6 |
a45915bc778c3b967b94f7cee4faa46d | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to show the existence of acute triangles in $\{b\}.$ Note that $b_1 = 1$ and for all $i$ $b_i \le n.$
Now three arbitrary sidelengths $x$ $y$ , and $z$ , with $x \le y \le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$ 's for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$ ).
We now make another substitution: $c_i \equiv b_i ^2.$ So $c_1 = 1$ and for all $i$ $c_i \le n^2.$ Now we examine the smallest possible sets $\{c\}$ for small $n$ for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If $n=3$ , then the smallest possible set, call it $\{s_3\},$ is trivially $\{1,1,2\}$ , since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\{s_n\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\{s_n\} = \{F_0, F_1, ... F_n\}$ , then $\{s_{n+1}\} = \{F_0, F_1, ... F_n, c_{n+1}\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\{s_n\}$ which are $F_{n-1}$ and $F_n$ . But these sum to $F_{n+1}$ so $\{s_{n+1}\} = \{F_0, F_1, ... F_{n+1}\}$ and our induction is complete.
Now since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set $\{c\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\{c\}$ is bounded between $1$ and $n^2$ , then the conditions of the problem are met if and only if $F_{n-1} > n^2$ . The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\boxed{13}$ | null | 13 |
a45915bc778c3b967b94f7cee4faa46d | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Outline:
1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \ge 3$
2. If the chosen $n$ is such that $F_n \le n^2$ , then choose the sequence $a_n$ such that $a_k = \sqrt{F_k}$ for $1 \le k \le n$ . It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to $n$ times the smallest term. Also, because for any three terms $x = \sqrt{F_a}, y = \sqrt{F_b}, z = \sqrt{F_c}$ with $a<b<c$ $x^2 + y^2 = F_a + F_b \le F_{b-1} + F_b = F_{b+1} \le F_c = z^2$ , x, y, z do not form an acute triangle. Thus, all $n$ such that $F_n \le n^2$ do not work.
3. It is easy to observe via a contradiction argument that all $n$ such that $F_n > n^2$ produce an acute triangle. (If, without loss of generality, $a_n$ is an increasing sequence, such that no three (in particular, consecutive) terms form an acute triangle, then $a_2^2 \ge F_1a_1^2, a_3^2 \ge a_2^2 + a_1^2 \ge F_2a^2$ , and by induction $a_n^2 > F_na_1^2$ , a contradiction to the condition's inequality.)
4. Note that $F_{12} = 144 = 12^2$ and $F_{13} = 233 > 169 = 13^2$ . It is easily to verify through strong induction that all $n$ greater than 12 make $F_n > n^2$ . Thus, $\boxed{13}$ is the desired solution set. | null | 13 |
d89b02c176f387e6b9c6a679272db18b | https://artofproblemsolving.com/wiki/index.php/2000_USAMO_Problems/Problem_4 | Find the smallest positive integer $n$ such that if $n$ squares of a $1000 \times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board. | We claim that $n = 1999$ is the smallest such number. For $n \le 1998$ , we can simply color any of the $1998$ squares forming the top row and the left column, but excluding the top left corner square.
We now show that no configuration with no colored right triangles exists for $n = 1999$ . We call a row or column filled if all $1000$ of its squares are colored. Then any of the remaining $999$ colored squares must share a column or row, respectively, with one of the colored squares in a filled row or column. These two squares, and any other square in the filled row or column, form a colored right triangle, giving us a contradiction. Hence, no filled row or column may exist.
Let $m$ be the number of columns with $1$ colored square. Then there are $1999-m$ colored squares in the remaining columns, and in each of these $< 1999-m$ columns that have a colored square must have at least two colored squares in them. These two colored squares will form a triangle with any other colored square in either of the rows containing the colored squares. Hence, each of the $1999-m$ colored squares must be placed in different rows, but as there are only $1000$ rows, the inequality $1999 - m \le 1000 \Longrightarrow m \ge 999$ holds. If $m = 1000$ , then each column only has $1$ colored square, leaving no place for the remaining $999$ , contradiction. If $m = 999$ , then each of the $1000$ rows has $1$ black square, leaving no place for the other $999$ , contradiction. Hence $n = \boxed{1999}$ is the minimal value. | null | 1999 |
2dec4c965bc88a3697fd054e8ba36687 | https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_1 | In triangle $ABC$ , angle $A$ is twice angle $B$ , angle $C$ is obtuse , and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter | [asy] import olympiad; pair A, B, C, D, extensionAC; real angleABC; path braceBC; A = (0, 0); B = (2, 0); D = (1, .5); angleABC = atan(.5); //y = 4x/3 and x+2y = 2 (sides AC and BC, respectively) intersect here: C = (6/11, 8/11); braceBC = brace(C, B, .1); label("$\mathsf{A}$", A, W); label("$\mathsf{B}$", B, E); label("$\mathsf{C}$", C, N); label("$\mathsf{D}$", D, S); label("$\mathsf{a}$", braceBC, NE); label("$\mathsf{b}$", A--C, NW); label("$\mathsf{c}$", A--B, S); label("$\mathsf{x}$", A--D, N); draw(A--B--C--cycle); draw(A--D); draw(anglemark(C, B, A)); draw(anglemark(B, A, D)); draw(anglemark(D, A, C)); draw(braceBC); [/asy] (diagram by integralarefun)
After drawing the triangle, also draw the angle bisector of $\angle A$ , and let it intersect $\overline{BC}$ at $D$ . Notice that $\triangle ADC\sim \triangle BAC$ , and let $AD=x$ . Now from similarity, \[x=\frac{bc}{a}\] However, from the angle bisector theorem, we have \[BD=\frac{ac}{b+c}\] but $\triangle ABD$ is isosceles, so \[x=BD\Longrightarrow \frac{bc}{a}=\frac{ac}{b+c}\Longrightarrow a^2=b(b+c)\] so all sets of side lengths which satisfy the conditions also meet the boxed condition.
Notice that $\text{gcd}(a, b, c)=1$ or else we can form a triangle by dividing $a, b, c$ by their greatest common divisor to get smaller integer side lengths, contradicting the perimeter minimality. Since $a$ is squared, $b$ must also be a square because if it isn't, then $b$ must share a common factor with $b+c$ , meaning it also shares a common factor with $c$ , which means $a, b, c$ share a common factor—a contradiction. Thus we let $b = x^2, b+c = y^2$ , so $a = xy$ , and we want the minimal pair $(x,y)$
By the Law of Cosines \[b^2 = a^2 + c^2 - 2ac\cos B\]
Substituting $a^2 = b^2 + bc$ yields $\cos B = \frac{b+c}{2a} = \frac{y}{2x}$ . Since $\angle C > 90^{\circ}$ $0^{\circ} < \angle B < 30^{\circ} \Longrightarrow \sqrt{3} < \frac{y}{x} < 2$ . For $x \le 3$ there are no integer solutions. For $x = 4$ , we have $y = 7$ that works, so the side lengths are $(a, b, c)=(28, 16, 33)$ and the minimal perimeter is $\boxed{77}$ | null | 77 |
2dec4c965bc88a3697fd054e8ba36687 | https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_1 | In triangle $ABC$ , angle $A$ is twice angle $B$ , angle $C$ is obtuse , and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter | In $\triangle ABC$ let $\angle B = \beta, \angle A = 2\beta, \angle C = 180^{\circ} - 3\beta$ . From the law of sines, we have \[\frac{a}{\sin 2\beta} = \frac{b}{\sin \beta} = \frac{c} {\sin (180^{\circ} - 3\beta)} = \frac{c}{\sin 3\beta}\] Thus the ratio \[b : a : c = \sin\beta : \sin 2\beta : \sin 3\beta\] We can simplify \[\frac{\sin 2\beta}{\sin\beta} = \frac{2\sin\beta\cos\beta}{\sin\beta} = 2\cos\beta\] Likewise, \[\frac{\sin 3\beta}{\sin\beta} = \frac{\sin 2\beta\cos\beta + \sin\beta\cos 2\beta}{\sin\beta} = \frac{2\sin\beta\cos^2\beta + \sin\beta(\cos^2\beta - \sin^2\beta)}{\sin\beta}\] \[= {2 \cos^2 \beta + \cos^2 \beta - \sin^2 \beta} = 4\cos^2 \beta - 1\] Letting $\gamma = \cos\beta$ , rewrite \[b : a : c = 1 : 2\gamma : 4\gamma^2 - 1\]
We find that to satisfy the conditions for an obtuse triangle, $\beta \in (0^\circ, 30^\circ)$ and therefore $\gamma \in \left(\frac{\sqrt{3}}{2}, 1\right)$
The rational number with minimum denominator (in order to minimize scaling to obtain integer solutions) above $\frac{\sqrt{3}}{2}$ is $\frac{7}{8}$ , which also has a denominator divisible by 2 (to take advantage of the coefficients of 2 and 4 in the ratio and further minimize scaling).
Inserting $\gamma = \frac{7}{8}$ into the ratio, we find $b : a : c = 1 : \frac{7}{4} : \frac{33}{16}$ . When scaled minimally to obtain integer side lengths, we find \[b, a, c = 16, 28, 33\] and that the perimeter is $\boxed{77}$ | null | 77 |
bc9fb059b7f4baf332344340e897b028 | https://artofproblemsolving.com/wiki/index.php/1986_USAMO_Problems/Problem_3 | What is the smallest integer $n$ , greater than one, for which the root-mean-square of the first $n$ positive integers is an integer?
$\mathbf{Note.}$ The root-mean-square of $n$ numbers $a_1, a_2, \cdots, a_n$ is defined to be \[\left[\frac{a_1^2 + a_2^2 + \cdots + a_n^2}n\right]^{1/2}\] | Let's first obtain an algebraic expression for the root mean square of the first $n$ integers, which we denote $I_n$ . By repeatedly using the identity $(x+1)^3 = x^3 + 3x^2 + 3x + 1$ , we can write \[1^3 + 3\cdot 1^2 + 3 \cdot 1 + 1 = 2^3,\] \[1^3 + 3 \cdot(1^2 + 2^2) + 3 \cdot (1 + 2) + 1 + 1 = 3^3,\] and \[1^3 + 3\cdot(1^2 + 2^2 + 3^2) + 3 \cdot (1 + 2 + 3) + 1 + 1 + 1 = 4^3.\] We can continue this pattern indefinitely, and thus for any
positive integer $n$ \[1 + 3\sum_{j=1}^n j^2 + 3 \sum_{j=1}^n j^1 + \sum_{j=1}^n j^0 = (n+1)^3.\] Since $\sum_{j=1}^n j = n(n+1)/2$ , we obtain \[\sum_{j=1}^n j^2 = \frac{2n^3 + 3n^2 + n}{6}.\] Therefore, \[I_n = \left(\frac{1}{n} \sum_{j=1}^n j^2\right)^{1/2} = \left(\frac{2n^2 + 3n + 1}{6}\right)^{1/2}.\] Requiring that $I_n$ be an integer, we find that \[(2n+1 ) (n+1) = 6k^2,\] where $k$ is an integer. Using the Euclidean algorithm, we see that $\gcd(2n+1, n+1) = \gcd(n+1,n) = 1$ , and so $2n+1$ and $n+1$ share no
factors greater than 1. The equation above thus implies that $2n+1$ and $n+1$ is each proportional to a perfect square. Since $2n+1$ is
odd, there are only two possible cases:
Case 1: $2n+1 = 3 a^2$ and $n+1 = 2b^2$ , where $a$ and $b$ are integers.
Case 2: $2n+1 = a^2$ and $n+1 = 6b^2$
In Case 1, $2n+1 = 4b^2 -1 = 3a^2$ . This means that $(4b^2 -1)/3 = a^2$ for some integers $a$ and $b$ . We proceed by checking whether $(4b^2-1)/3$ is a perfect square for $b=2, 3, 4, \dots$ . (The solution $b=1$ leads to $n=1$ , and we are asked to find a value of $n$ greater than 1.) The smallest positive integer $b$ greater than 1 for
which $(4b^2-1)/3$ is a perfect square is $b=13$ , which results in $n=337$
In Case 2, $2n+1 = 12b^2 - 1 = a^2$ . Note that $a^2 = 2n+1$ is an odd square, and hence is congruent to $1 \pmod 4$ . But $12b^2 -1 \equiv 3 \pmod 4$ for any $b$ , so Case 2 has no solutions.
Alternatively, one can proceed by checking whether $12b^2 -1$ is a perfect square for $b=1, 2 ,3 ,\dots$ . We find that $12b^2 -1$ is not a perfect square for $b = 1,2, 3, ..., 7, 8$ , and $n= 383$ when $b=8$ . Thus the smallest positive integers $a$ and $b$ for which $12b^2- 1 = a^2$ result in a value of $n$ exceeding the value found in Case 1, which was 337.
In summary, the smallest value of $n$ greater than 1 for which $I_n$ is an integer is $\boxed{337}$ | null | 337 |
982766bb6c57b0aec7d8441644d076c0 | https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_5 | Let $a_1,a_2,a_3,\cdots$ be a non-decreasing sequence of positive integers. For $m\ge1$ , define $b_m=\min\{n: a_n \ge m\}$ , that is, $b_m$ is the minimum value of $n$ such that $a_n\ge m$ . If $a_{19}=85$ , determine the maximum value of $a_1+a_2+\cdots+a_{19}+b_1+b_2+\cdots+b_{85}$ | We create an array of dots like so: the array shall go out infinitely to the right and downwards, and at the top of the $i$ th column we fill the first $a_i$ cells with one dot each. Then the $19$ th row shall have 85 dots. Now consider the first 19 columns of this array, and consider the first 85 rows. In row $j$ , we see that the number of blank cells is equal to $b_j-1$ . Therefore the number of filled cells in the first 19 columns of row $j$ is equal to $20-b_j$
We now count the number of cells in the first 19 columns of our array, but we do it in two different ways. First, we can sum the number of dots in each column: this is simply $a_1+\cdots+a_{19}$ . Alternatively, we can sum the number of dots in each row: this is $(20-b_1)+\cdots +(20-b_{85})$ . Since we have counted the same number in two different ways, these two sums must be equal. Therefore \[a_1+\cdots +a_{19}+b_1+\cdots +b_{85}=20\cdot 85=\boxed{1700}.\] Note that this shows that the value of the desired sum is constant. | null | 1700 |
adbce7149b71691685e6ff8865546ef9 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | Using Vieta's formulas, we have:
\begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*}
From the last of these equations, we see that $cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62$ . Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$ , and so $ac+ad+bc+bd=k-30$ . The key insight is now to factor the left-hand side as a product of two binomials: $(a+b)(c+d)=k-30$ , so that we now only need to determine $a+b$ and $c+d$ rather than all four of $a,b,c,d$
Let $p=a+b$ and $q=c+d$ . Plugging our known values for $ab$ and $cd$ into the third Vieta equation, $-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)$ , we have $-200 = -32(c+d) + 62(a+b) = 62p-32q$ . Moreover, the first Vieta equation, $a+b+c+d=18$ , gives $p+q=18$ . Thus we have two linear equations in $p$ and $q$ , which we solve to obtain $p=4$ and $q=14$
Therefore, we have $(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30$ , yielding $k=4\cdot 14+30 = \boxed{86}$ | null | 86 |
adbce7149b71691685e6ff8865546ef9 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | We start as before: $ab=-32$ and $cd=62$ . We now observe that a and b must be the roots of a quadratic, $x^2+rx-32$ , where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic $x^2+sx+62$
Now
\begin{align*}x^4-18x^3+kx^2+200x-1984 =& (x^2+rx-32)(x^2+sx+62)\\ =& x^4+(r+s)x^3+(62-32+rs)x^2\\ &+(62s-32r)x-1984.\end{align*}
Equating the coefficients of $x^3$ and $x$ with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of $x^2$ and get $k=\boxed{86}.$ | null | 86 |
adbce7149b71691685e6ff8865546ef9 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | Let the roots of the equation be $a,b,c,$ and $d$ . By Vieta's, \begin{align*} a+b+c+d &= 18\\ ab+ac+ad+bc+bd+cd &= k\\ abc+abd+acd+bcd &=-200\\ abcd &=-1984.\\ \end{align*} Since $abcd=-1984$ and $ab=-32$ , then, $cd=62$ . Notice that \[abc + abd + acd + bcd = -200\] can be factored into \[ab(c+d)+cd(a+b)=-32(c+d)+62(a+b).\] From the first equation, $c+d=18-a-b$ . Substituting it back into the equation, \[-32(18-a-b)+62(a+b)=-200\] Expanding, \[-576+32a+32b+62a+62b=-200 \implies 94a+94b=376\] So, $a+b=4$ and $c+d=14$ . Notice that \[ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)\] Plugging all our values in, \[-32+62+4(14)=\boxed{86}.\] | null | 86 |
adbce7149b71691685e6ff8865546ef9 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | Since two of the roots have product $-32,$ the equation can be factored in the form \[x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + c).\] Expanding, we get \[x^4 - 18x^3 + kx^2 + 200x - 1984 = x^4 + (a + b) x^3 + (ab + c - 32) x^2 + (ac - 32b) x - 32c = 0.\] Matching coefficients, we get
\begin{align*}
a + b &= -18, \\
ab + c - 32 &= k, \\
ac - 32b &= 200, \\
-32c &= -1984.
\end{align*}Then $c = \frac{-1984}{-32} = 62,$ so $62a - 32b = 200.$ With $a + b = -18,$ we can solve to find $a = -4$ and $b = -14.$ Then \[k = ab + c - 32 = \boxed{86}.\] | null | 86 |
ee0e8c72a8aeb1e1cbd508babe7d886e | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_1 | In a party with $1982$ people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else? | We induct on $n$ to prove that in a party with $n$ people, there must be at least $(n-3)$ people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people $A, B, C$ , who do not know each other.)
Base case: $n = 4$ is obvious.
Inductive step: Suppose in a party with $k$ people (with $k \ge 4$ ), at least $(k-3)$ people know everyone else. Consider a party with $(k+1)$ people. Take $k$ of the people (leaving another person, $A$ , out) and apply the inductive step to conclude that at least $(k-3)$ people know everyone else in the $k$ -person group, $G$
Now suppose that everyone in the group $G$ knows each other. Then take $3$ of these people and $A$ to deduce that $A$ knows a person $B \in G$ , which means $B$ knows everyone else. Then apply the inductive step on the remaining $k$ people (excluding $B$ ) to find $(k-3)$ people out of them that know everyone else (including $B$ , of course). Then these $(k-3)$ people and $B$ , which enumerate $(k-2)$ people, know everyone else.
Suppose that there exist two people $B, C \in G$ who do not know each other. Because $k-3 \ge 1$ , there exist at least one person in $G$ , person $D$ , who knows everyone else in $G$ . Now, take $A, B, C, D$ and observe that because $B, C$ do not know each other, either $A$ or $D$ knows everyone else of $A, B, C, D$ (by the problem condition), so in particular $A$ and $D$ know each other. Then apply the inductive step on the remaining $k$ people (excluding $D$ ) to find $(k-3)$ people out of them that know everyone else (including $D$ , of course). Then these $(k-3)$ people and $D$ , which enumerate $(k-2)$ people, know everyone else.
This completes the inductive step and thus the proof of this stronger result, which easily implies that at least $1982 - 3 = \boxed{1979}$ people know everyone else. | null | 1979 |
c4b2a2c596bc5c3b88aec10541b75586 | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_2 | Let $S_r=x^r+y^r+z^r$ with $x,y,z$ real. It is known that if $S_1=0$
$(*)$ $\frac{S_{m+n}}{m+n}=\frac{S_m}{m}\frac{S_n}{n}$
for $(m,n)=(2,3),(3,2),(2,5)$ , or $(5,2)$ . Determine all other pairs of integers $(m,n)$ if any, so that $(*)$ holds for all real numbers $x,y,z$ such that $x+y+z=0$ | Claim Both $m,n$ can not be even.
Proof $x+y+z=0$ $\implies x=-(y+z)$
Since $\frac{S_{m+n}}{m+n} = \frac{S_m S_n}{mn}$
by equating cofficient of $y^{m+n}$ on LHS and RHS ,get
$\frac{2}{m+n}=\frac{4}{mn}$
$\implies \frac{m}{2} + \frac {n}{2} = \frac{m\cdot n}{2\cdot2}$
So we have, $\frac{m}{2} \biggm{|} \frac{n}{2}$ and $\frac{n}{2} \biggm{|} \frac{m}{2}$
$\implies m=n=4$
So we have $S_8=2(S_4)^2$
Now since it will true for all real $x,y,z,x+y+z=0$ .
So choose $x=1,y=-1,z=0$
$S_8=2$ and $S_4=2$ so $S_8 \neq 2 S_4^2$
This is contradiction. So, at least one of $m,n$ must be odd. WLOG assume $n$ is odd and m is even. The coefficient of $y^{m+n-1}$ in $\frac{S_{m+n}}{m+n}$ is $\frac{\binom{m+n}{1} }{m+n} =1$
The coefficient of $y^{m+n-1}$ in $\frac{S_m\cdot S_n}{m\cdot n}$ is $\frac{2}{m}$
Therefore, $\boxed{2}$ | null | 2 |
12319197a4824f2c1f825b93a283e3bb | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_4 | Prove that there exists a positive integer $k$ such that $k\cdot2^n+1$ is composite for every integer $n$ | Indeed, $\boxed{2935363331541925531}$ has the requisite property. | null | 2935363331541925531 |
02c5b310832b5c75b71a805a7f2ad412 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_4 | Determine all the roots real or complex , of the system of simultaneous equations | Let $P(t)=t^3-at^2+bt-c$ have roots x, y, and z. Then \[0=P(x)+P(y)+P(z)=3-3a+3b-3c\] using our system of equations, so $P(1)=0$ . Thus, at least one of x, y, and z is equal to 1; without loss of generality, let $x=1$ . Then we can use the system of equations to find that $y=z=1$ as well, and so $\boxed{1,1,1}$ is the only solution to the system of equations. | null | 1,1,1 |
02c5b310832b5c75b71a805a7f2ad412 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_4 | Determine all the roots real or complex , of the system of simultaneous equations | Let $a=x-1,$ $b=y-1$ and $c=z-1.$ Then \[a+b+c=0,\] \[a^2+b^2+c^2=0,\] \[a^3+b^3+c^3=0.\] We have \begin{align*} 0&=(a+b+c)^3\\ &=(a^3+b^3+c^3)+3a^2(b+c)+3b^3(a+c)+3c^2(a+b)+6abc\\ &=0-3a^3-3b^3-3c^3+6abc\\ &=6abc. \end{align*} Then one of $a, b$ and $c$ has to be 0, and easy to prove the other two are also 0. So $\boxed{1,1,1}$ is the only solution to the system of equations. | null | 1,1,1 |
02c5b310832b5c75b71a805a7f2ad412 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_4 | Determine all the roots real or complex , of the system of simultaneous equations | We are going to use Intermediate Algebra Techniques to solve this equation.
Let's start with the first one: $x+y+z=3$ . This will be referred as the FIRST equation.
We are going to use the first equation to relate to the SECOND one ( $x^2+y^2+z^2=3$ ) and the THIRD one ( $x^3+y^3+z^3=3)$
Squaring this equation: $x^2+y^2+z^2+2xy+2yz+2xz=9$
Subtracting this equation from the 2nd equation in the problem, we have $2xy+2yz+2xz=6$ , so $xy+xz+yz=3$
Now we try the same idea with the cubed terms. Cube the first equation:
$x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3yz^2+3zx^2+3zy^2=27$ . Plug in $x^3+y^3+z^3=3$ and factor partially:
$3+3(x^2(y+z)+y^2(x+z)+z^2(x+y))+6xyz=27$
Now here is the key step. Note that $z=3-x-y, y=3-x-z, x=3-y-z$ . So we are going to substitute $y+z, x+z, x+y$ for each of the expressions and we get: $-x^3+3x^2-y^3+3y^2-z^3+3z^2=8-2xyz$ (I rearranged it a bit).
Resubstituting in the second and third equation: $-3+3(3)=8-2xyz$ . So $xyz=1$
So now we have three equations for the elementary symmetric sums of $x,y,z$
Equation 4: $x+y+z=3$ (this is also equation 1)
Equation 5: $xy+yz+xz=3$
Equation 6: $xyz=1$
If we call the solutions of $t^3-3t^2+3t-1=0$ (Equation 7) $a,b,c$ , then $x,y,z$ are the three roots $a,b,c$ but in some order. Notice that Equation 7 can be factored as $(t-1)^3=0$ , which means that $t=1$ . Therefore $(x,y,z)$ are permutations of $(1,1,1)$ in some order, which can be only $(1,1,1)$ . (This step uses Vieta's formulas)
Therefore, the only solution is $\boxed{1,1,1}$ | null | 1,1,1 |
91064cf7717d0f5be97285821d1c166e | https://artofproblemsolving.com/wiki/index.php/2021_USAJMO_Problems/Problem_1 | Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\] | The answer is $\boxed{1}$ , which works.
To show it is necessary, we first get $f(1)=f(1)^2$ , so $f(1)=1$ .
Then, we get $f(2)=f(1^2 + 1^2)=f(1)^2 =1$ | null | 1 |
5878a9d27bdabf85aea862589abc464f | https://artofproblemsolving.com/wiki/index.php/2020_USOJMO_Problems/Problem_3 | An empty $2020 \times 2020 \times 2020$ cube is given, and a $2020 \times 2020$ grid of square unit cells is drawn on each of its six faces. A beam is a $1 \times 1 \times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:
What is the smallest positive number of beams that can be placed to satisfy these conditions? | Place the cube in the xyz-coordinate, with the positive x-axis pointing forward, the positive y-axis pointing right, and the positive z-axis pointing up. Let the position of a unit cube be $(x, y, z)$ if it is centered at $(x, y, z)$ . Place the $2020 \times 2020 \times 2020$ cube so that the edges are parallel to the axes, and two of its corners are at $(1, 1, 1)$ and $(2020, 2020, 2020)$ . Now call a beam z-oriented if its endpoints differ only in z-coordinates, and similarly call it x-oriented or y-oriented if the endpoints differ in x- or y-coordinates, respectively.
We claim that the answer is $\boxed{3030}$ . First we will prove 3030 suffices. Place the first beam so its endpoints lie at $(1, 1, 1)$ , and $(2020, 1, 1)$ . Place the 2nd beam so its endpoints lie at $(1, 1, 2)$ and $(1, 2020, 2)$ . Place the 3rd beam so its endpoints lie at $(2, 2, 1)$ and $(2, 2, 2020)$ . Now it is clear that the only faces among these three beams that are not touching the faces of the $2020 \times 2020 \times 2020$ cube of the face of another beam are the top face of beam 2, and the front and right faces of beam 3. | null | 3030 |
da6969b6e725522504576dde9744f9cc | https://artofproblemsolving.com/wiki/index.php/2020_USOJMO_Problems/Problem_5 | Suppose that $(a_1,b_1),$ $(a_2,b_2),$ $\dots,$ $(a_{100},b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i,j)$ satisfying $1\leq i<j\leq 100$ and $|a_ib_j-a_jb_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs. | Call the pair $(i, j)$ good if $1\leq i < j \leq 100$ and $|a_ib_j-a_jb_i|=1$ . Note that we can reorder the pairs $(a_1, b_1), (a_2, b_2), \ldots, (a_{100}, b_{100})$ without changing the number of good pairs. Thus, we can reorder them so that $a_1\leq a_2\leq\ldots\leq a_{100}$ . Furthermore, reorder them so that if $a_i=a_j$ for some $i<j$ , then $b_i<b_j$
Now I claim the maximum value of $N$ is $\boxed{197}$ . First, we will show $N\leq 197$ | null | 197 |
8f163c68b8bf11a7dbb1d1ef054ef7cc | https://artofproblemsolving.com/wiki/index.php/2016_USAJMO_Problems/Problem_4 | Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,...,N\}$ , one can still find $2016$ distinct numbers among the remaining elements with sum $N$ | Since any $2016$ elements are removed, suppose we remove the integers from $1$ to $2016$ . Then the smallest possible sum of $2016$ of the remaining elements is \[2017+2018+\cdots + 4032 = 1008 \cdot 6049 = 6097392\] so clearly $N\ge 6097392$ . We will show that $N=6097392$ works.
$\vspace{0.2 in}$
$\{1,2\cdots 6097392\}$ contain the integers from $1$ to $6048$ , so pair these numbers as follows:
\[1, 6048\]
\[2, 6047\]
\[3, 6046\]
\[\cdots\]
\[3024, 3025\]
When we remove any $2016$ integers from the set $\{1,2,\cdots N\}$ , clearly we can remove numbers from at most $2016$ of the $3024$ pairs above, leaving at least $1008$ complete pairs. To get a sum of $N$ , simply take these $1008$ pairs, all of which sum to $6049$ . The sum of these $2016$ elements is $1008 \cdot 6049 = 6097392$ , as desired.
$\vspace{0.2 in}$
We have shown that $N$ must be at least $6097392$ , and that this value is attainable. Therefore our answer is $\boxed{6097392}$ | null | 6097392 |
f25c446de036ffaadaf5676a0b0756b1 | https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_4 | problem_id
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
Name: Text, dtype: object | We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\boxed{6}$ | null | 6 |
b8db09096d7162ee8404b842152eba4b | https://artofproblemsolving.com/wiki/index.php/2013_USAJMO_Problems/Problem_4 | Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$ $2+2$ $2+1+1$ $1+2+1$ $1+1+2$ , and $1+1+1+1$ . Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd. | First of all, note that $f(n)$ $\sum_{i=0}^{k} f(n-2^{i})$ where $k$ is the largest integer such that $2^k \le n$ . We let $f(0) = 1$ for convenience.
From here, we proceed by induction, with our claim being that the only $n$ such that $f(n)$ is odd are $n$ representable of the form $2^{a} - 1, a \in \mathbb{Z}$
We induct on $a$ . It is trivially true for $a = 0$ and $a = 1$ . From here, we show that, if the only numbers $n \le 2^{a-1} - 1$ where $f(n)$ is odd are of the form described above, then the only numbers $n \le 2^{a} -1$ that are odd are of that form. We first consider all numbers $b$ , such that $2^{a-1} \le b \le 2^{a} - 2$ , going from the lower bound to the upper bound (a mini induction, you might say). We know that $f(b) = \sum_{i=0}^{a-1} f(b-2^{i})$ . For a number in this summation to be odd, $b - 2^i = 2^m -1 \rightarrow b = 2^i + 2^m - 1$ . However, we know that $b > 2^{a-1}$ , so $m$ must be equal to $a-1$ , or else $b$ cannot be in that interval. Now, from this, we know that $i < a-1$ , as $b<2^{a} - 1$ . Therefore, $i$ and $m$ are distinct, and thus $f(b - 2^i)$ and $f(b- 2^{a-1})$ are odd; since there are just two odd numbers, the ending sum for any $b$ is even. Finally, considering $2^{a} - 1$ , the only odd number is $f(2^{a} - 1 - 2^{a-1})$ , so the ending sum is odd. $\Box$
The smallest $n$ greater than $2013$ expressible as $2^d - 1, d \in \mathbb{N}$ is $2^{11} -1 = \boxed{2047}$ | null | 2047 |
b8db09096d7162ee8404b842152eba4b | https://artofproblemsolving.com/wiki/index.php/2013_USAJMO_Problems/Problem_4 | Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$ $2+2$ $2+1+1$ $1+2+1$ $1+1+2$ , and $1+1+1+1$ . Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd. | Of course, as with any number theory problem, use actual numbers to start, not variables! By plotting out the first few sums (do it!) and looking for patterns, we observe that $f(n)=\sum_{\textrm{power}=0}^{\textrm{pow}_{\textrm{larg}}} f(n-2^{\textrm{power}})$ , where $\textrm{pow}_{\textrm{larg}}$ represents the largest power of $2$ that is smaller than $n$ . I will call this sum the Divine Sign, or DS.
But wait a minute... we are trying to determine odd/even of $f(n)$ . Why not call all the evens 0 and odds 1, basically using mod 2? Sounds so simple. Draw a small table for the values: as $n$ goes up from $0$ , you get: $1,1,0,1,0,0,0,1,0...$ . We have to set $f(0)=1$ for this to work. Already it looks like $f(n)$ is only odd if $n=2^{\textrm{power}}-1$
The only tool here is induction. The base case is clearly established. Then let's assume we successfully made our claim up to $2^n-1$ . We need to visit numbers from $2^n$ to $2^{n+1}-1$ . Realize that $2^n$ has $0$ for $f$ because there will be two numbers in DS that give a $f$ of one: $2^{n-1}$ and $1$
But to look at whether a value of $f(\textrm{number})$ is 1 or 0, we need to revisit our first equation. We can answer this rather natural question: When will a number to be inducted upon, say $2^n+k$ , ever have a 1 as $f(\textrm{number})$ in the DS equation? Well- because by our assumption of the claim up to $2^n-1$ , we know that the only way for that to happen is if $2^n+k-2^{\textrm{power}}$ in the DS is equal to $2^{\textrm{Some power}} - 1$ . Clearly $1 \leq k \leq 2^n - 1$
Finally, we can simplify. Using our last equation, $2^n+k-2^{\textrm{power}}=2^{\textrm{Some power}}-1$ , regrouping gives $2^n+k=2^{\textrm{power}}+2^{\textrm{Some power}}-1$
Most importantly, realize that $\textrm{power}$ can be from $0$ to $n$ , because of the restraints on $k$ mentioned earlier. Same with $\textrm{Some power}$ . Immediately at least one of $\textrm{power}$ and $\textrm{Some power}$ has to be $n$ . If both were smaller, LHS is greater, contradiction. If both were greater, RHS is greater, contradiction.
Therefore, by setting one of $\textrm{power}$ or $\textrm{Some power}$ to $n$ , we realize $k=2^{\textrm{A certain power}}-1$
The conclusion is clear, right? Each $k$ from $1$ to $2^{n-1}-1$ yields two distinct cases: one of $\textrm{power}$ and $\textrm{Some power}$ is equal to $n$ , while the other is LESS THAN $n$ . But for $k=2^n-1$ , there is ONE CASE: BOTH values have to equal $n$ . Therefore, the only $k$ that has $f(2^n+k)$ as odd must only be $2^n-1$ , because the other ones yield a $f$ of 1+1=0 in our mod. That proves our induction for a new power of 2, namely $n+1$ , meaning that $f(\textrm{number})$ is only odd if $\textrm{number} = 2^{\textrm{Power of two}} - 1$ , and we are almost done...
Thus, the answer is $2^{11}-1=\boxed{2047}$ | null | 2047 |
a45915bc778c3b967b94f7cee4faa46d | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to show the existence of acute triangles in $\{b\}.$ Note that $b_1 = 1$ and for all $i$ $b_i \le n.$
Now three arbitrary sidelengths $x$ $y$ , and $z$ , with $x \le y \le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$ 's for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$ ).
We now make another substitution: $c_i \equiv b_i ^2.$ So $c_1 = 1$ and for all $i$ $c_i \le n^2.$ Now we examine the smallest possible sets $\{c\}$ for small $n$ for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If $n=3$ , then the smallest possible set, call it $\{s_3\},$ is trivially $\{1,1,2\}$ , since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\{s_n\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\{s_n\} = \{F_0, F_1, ... F_n\}$ , then $\{s_{n+1}\} = \{F_0, F_1, ... F_n, c_{n+1}\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\{s_n\}$ which are $F_{n-1}$ and $F_n$ . But these sum to $F_{n+1}$ so $\{s_{n+1}\} = \{F_0, F_1, ... F_{n+1}\}$ and our induction is complete.
Now since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set $\{c\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\{c\}$ is bounded between $1$ and $n^2$ , then the conditions of the problem are met if and only if $F_{n-1} > n^2$ . The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\boxed{13}$ | null | 13 |
a45915bc778c3b967b94f7cee4faa46d | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Outline:
1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \ge 3$
2. If the chosen $n$ is such that $F_n \le n^2$ , then choose the sequence $a_n$ such that $a_k = \sqrt{F_k}$ for $1 \le k \le n$ . It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to $n$ times the smallest term. Also, because for any three terms $x = \sqrt{F_a}, y = \sqrt{F_b}, z = \sqrt{F_c}$ with $a<b<c$ $x^2 + y^2 = F_a + F_b \le F_{b-1} + F_b = F_{b+1} \le F_c = z^2$ , x, y, z do not form an acute triangle. Thus, all $n$ such that $F_n \le n^2$ do not work.
3. It is easy to observe via a contradiction argument that all $n$ such that $F_n > n^2$ produce an acute triangle. (If, without loss of generality, $a_n$ is an increasing sequence, such that no three (in particular, consecutive) terms form an acute triangle, then $a_2^2 \ge F_1a_1^2, a_3^2 \ge a_2^2 + a_1^2 \ge F_2a^2$ , and by induction $a_n^2 > F_na_1^2$ , a contradiction to the condition's inequality.)
4. Note that $F_{12} = 144 = 12^2$ and $F_{13} = 233 > 169 = 13^2$ . It is easily to verify through strong induction that all $n$ greater than 12 make $F_n > n^2$ . Thus, $\boxed{13}$ is the desired solution set. | null | 13 |
981346a8c4f46a2e3ea8c0d9d00c7ec5 | https://artofproblemsolving.com/wiki/index.php/2012_USAJMO_Problems/Problem_5 | For distinct positive integers $a$ $b < 2012$ , define $f(a,b)$ to be the number of integers $k$ with $1 \le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012. Let $S$ be the minimum value of $f(a,b)$ , where $a$ and $b$ range over all pairs of distinct positive integers less than 2012. Determine $S$ | Let $ak \equiv r_{a} \pmod{2012}$ and $bk \equiv r_{b} \pmod{2012}$ . Notice that this means $a(2012 - k) \equiv 2012 - r_{a} \pmod{2012}$ and $b(2012 - k) \equiv 2012 - r_{b} \pmod{2012}$ . Thus, for every value of $k$ where $r_{a} > r_{b}$ , there is a value of $k$ where $r_{b} > r_{a}$ . Therefore, we merely have to calculate $\frac{1}{2}$ times the number of values of $k$ for which $r_{a} \neq r_{b}$ and $r_{a} \neq 0$
However, the answer is NOT $\frac{1}{2}(2012) = 1006$ ! This is because we must count the cases where the value of $k$ makes $r_{a} = r_{b}$ or where $r_{a} = 0$
So, let's start counting.
If $k$ is even, we have either $a \equiv 0 \pmod{1006}$ or $a - b \equiv 0 \pmod{1006}$ . So, $a = 1006$ or $a = b + 1006$ . We have $1005$ even values of $k$ (which is all the possible even values of $k$ , since the two above requirements don't put any bounds on $k$ at all).
If $k$ is odd, if $k = 503$ or $k = 503 \cdot 3$ , then $a \equiv 0 \pmod{4}$ or $a \equiv b \pmod{4}$ . Otherwise, $ak \equiv 0 \pmod{2012}$ or $ak \equiv bk \pmod{2012}$ , which is impossible to satisfy, given the domain $a, b < 2012$ . So, we have $2$ values of $k$
In total, we have $2 + 1005 = 1007$ values of $k$ which makes $r_{a} = r_{b}$ or $r_{a} = 0$ , so there are $2011 - 1007 = 1004$ values of $k$ for which $r_{a} \neq r_{b}$ and $r_{a} \neq 0$ . Thus, by our reasoning above, our solution is $\frac{1}{2} \cdot 1004 = \boxed{502}$ | null | 502 |
f628ab1e2295d16ab604e15d621af50b | https://artofproblemsolving.com/wiki/index.php/2011_USAJMO_Problems/Problem_1 | Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square. | We will first take the expression modulo $3$ . We get $2^n+12^n+2011^n \equiv -1^n+1^n \pmod 3$
Lemma 1: All perfect squares are equal to $0$ or $1$ modulo $3$ .
We can prove this by testing the residues modulo $3$ . We have $0^2 \equiv 0 \pmod 3$ $1^2 \equiv 1 \pmod 3$ , and $2^2 \equiv 1 \pmod 3$ , so the lemma is true.
We know that if $n$ is odd, $-1^n+1^n \equiv 0 \pmod 3$ , which satisfies the lemma's conditions. However, if $n$ is even, we get $2 \pmod 3$ , which does not satisfy the lemma's conditions. So, we can conclude that $n$ is odd.
Now, we take the original expression modulo $4$ . For right now, we will assume that $n>1$ , and test $n=1$ later. For $n>1$ $2^n \equiv 0 \pmod 4$ , so $2^n+12^n+2011^n=-1^n \pmod 4$
Lemma 2: All perfect squares are equal to $0$ or $1$ modulo $4$ .
We can prove this by testing the residues modulo $4$ . We have $0^2 \equiv 0 \pmod 4$ $1^2 \equiv 1 \pmod 4$ $2^2 \equiv 0 \pmod 4$ , and $3^2 \equiv 1 \pmod 4$ , so the lemma is true.
We know that if $n$ is even, $-1^n \equiv 0 \pmod 4$ , which satisfies the lemma's conditions. However, if $n$ is odd, $-1^n \equiv -1 \equiv 3 \pmod 4$ , which does not satisfy the lemma's conditions. Therefore, $n$ must be even.
However, a number cannot be even and odd at the same time, so this is impossible. Now, we only have to test $n=1$ . We know that $2^1+12^1+2011^1=45^2$ , so the only integer is $\boxed{1}$ | null | 1 |
f9c3533133e74a99960a349b5036838f | https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1 | A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $ka_k$ is a perfect square for all $1\leq k\leq n$ . Find with proof the smallest $n$ such that $P(n)$ is a multiple of $2010$ | We claim that the smallest $n$ is $67^2 = \boxed{4489}$ | null | 4489 |
f9c3533133e74a99960a349b5036838f | https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1 | A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $ka_k$ is a perfect square for all $1\leq k\leq n$ . Find with proof the smallest $n$ such that $P(n)$ is a multiple of $2010$ | This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as "equivalence relation":
It is possible to write all positive integers $n$ in the form $p\cdot m^2$ , where $m^2$ is the largest perfect square dividing $n$ , so $p$ is not divisible by the square of any prime. Obviously, one working permutation of $[n]$ is simply $(1, 2, \ldots, n)$ ; this is acceptable, as $ka_k$ is always $k^2$ in this sequence.
Lemma 1. We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities $p$
Proof. Let $p_k$ and $m_k$ be the values of $p$ and $m$ , respectively, for a given $k$ as defined above, such that $p$ is not divisible by the square of any prime. We can obviously permute two numbers which have the same $p$ , since if $p_j = p_w$ where $j$ and $w$ are 2 values of $k$ , then $j\cdot w = p_j^2\cdot m_j^2\cdot m_w^2$ , which is a perfect square. This proves that we can permute any numbers with the same value of $p$
End Lemma
Lemma 2. We will prove the converse of Lemma 1: Let one number have a $p$ value of $\phi$ and another, $\gamma$ $\phi\cdot f$ and $\gamma\cdot g$ are both perfect squares.
Proof. $\phi\cdot f$ and $\gamma\cdot g$ are both perfect squares, so for $\phi\cdot \gamma$ to be a perfect square, if $g$ is greater than or equal to $f$ $g/f$ must be a perfect square, too. Thus $g$ is $f$ times a square, but $g$ cannot divide any squares besides $1$ , so $g = 1f$ $g = f$ . Similarly, if $f\geq g$ , then $f = g$ for our rules to keep working.
End Lemma
We can permute $l$ numbers with the same $p$ in $l!$ ways. We must have at least 67 numbers with a certain $p$ so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as $h$ , in general, we need numbers all the way up to $h\cdot 67^2$ , so obviously, $67^2$ is the smallest such number such that we can get a $67!$ term; here 67 $p$ terms are 1. Thus we need the integers $1, 2, \ldots, 67^2$ , so $67^2$ , or $\boxed{4489}$ , is the answer. | null | 4489 |
f9c3533133e74a99960a349b5036838f | https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1 | A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $ka_k$ is a perfect square for all $1\leq k\leq n$ . Find with proof the smallest $n$ such that $P(n)$ is a multiple of $2010$ | It's well known that there exists $f(n)$ and $g(n)$ such that $n = f(n) \cdot g(n)$ , no square divides $f(n)$ other than 1, and $g(n)$ is a perfect square.
We prove first: If $f(k) = f(a_k)$ $k \cdot a_k$ is a perfect square.
$k \cdot a_k = f(k) \cdot g(k) \cdot f(a_k) \cdot g(a_k) = f(k)^2 \cdot g(k) \cdot g(a_k)$ , which is a perfect square.
We will now prove: If $k \cdot a_k$ is a perfect square, $f(k) = f(a_k)$
We do proof by contrapositive: If $f(k) \neq f(a_k)$ $k \cdot a_k$ is not a perfect square.
$v_p(k)$ is the p-adic valuation of k. (Basically how many factors of p you can take out of k)
Note that if $f(k) \neq f(a_k)$ , By the Fundamental Theorem of Arithmetic, $f(k)$ and $f(a_k)$ 's prime factorization are different, and thus there exists a prime p, such that $v_p(f(k)) \neq v_p(f(a_k))$ . Also, since $f(k)$ and $f(a_k)$ is squarefree, $v_p(k), v_p(a_k) \leq 1$ . Thus, $v_p(k \cdot a_k) = 1$ , making $k \cdot a_k$ not a square.
Thus, we can only match k with $a_k$ if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the $a_k$ with f value 1, then 2, ... Thus, our answer is:
$P(n) = \prod_{1 \leq i \leq n, g(i) = 1} \left\lfloor \sqrt{\frac{n}{i}} \right \rfloor !$
For all $n < 67^2$ $P(n)$ doesn't have a factor of 67. However, if $n = 67^2$ , the first term will be a multiple of 2010, and thus the answer is $67^2 = \boxed{4489}$ | null | 4489 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \[12x+18x=45\] Solving gives us $x=1.5$ . The $18x$ is Alicia so $18\times1.5=\boxed{27}$ | E | 27 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | The relative speed of the two is $18+12=30$ , so $\frac{3}{2}$ hours would be required to travel $45$ miles. $d=st$ , so $x=18\cdot\frac{3}{2}=\boxed{27}$ | E | 27 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | Since $18$ mph is $\frac{3}{2}$ times $12$ mph, Alicia will travel $\frac{3}{2}$ times as far as Beth. If $x$ is the distance Beth travels, \[\frac{3}{2}x+x=45\] \[\frac{5}{2}x=45\] \[x=18\] Since this is the amount Beth traveled, the amount that Alicia traveled was \[45-18=\boxed{27}\] | E | 27 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | Alice and Barbara close in on each other at 30mph. Since they are 45 miles apart, they will meet in t = d/s = 45miles / 30mph = 3/2 hours. We can either calculate the distance Alice travels at 18mph or the distance Barbara travels at 12mph; since we want the distance from Alice, we go with the former. Alice (and Barbara) will meet in 1 1/2 hours at 18mph x 3/2 hours = 27 miles from A. $\boxed{27}$ | E | 27 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | We know that Beth starts 45 miles away from City A, let’s create two equations:
Alice-> $18t=d$ Beth-> $-12t+45=d$ [-12 is the slope; 45 is the y-intercept]
Solve the system:
$18t=-12t+45 30t=45 t=1.5$
So, $18(1.5)=$ $\boxed{27}$ | E | 27 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | Since Alicia and Beth's speeds are constant, they are directly proportional to their distances covered, so the ratio of their speeds is equal to the ratio of their covered distances. Since Alicia travels $\frac{18}{30} = \frac{3}{5}$ of their combined speed, she travels $\frac{3}{5}\cdot 45 = \boxed{27}$ of the total distance. | E | 27 |
a2658f3c6493f664125b0c5940db82a4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | Note that $40^2=1600$ but $45^{2}=2025$ (which is over our limit of $2023$ ). Therefore, the list is $5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2$ . There are $8$ elements, so the answer is $\boxed{8}$ | A | 8 |
a2658f3c6493f664125b0c5940db82a4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | Since $\left \lfloor{\sqrt{2023}}\right \rfloor = 44$ , there are $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{8}$ perfect squares less than 2023. | A | 8 |
a2658f3c6493f664125b0c5940db82a4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | Since $5$ is prime, each solution must be divisible by $5^2=25$ . We take $\left \lfloor{\frac{2023}{25}}\right \rfloor = 80$ and see that there are $\boxed{8}$ positive perfect squares no greater than $80$ | A | 8 |
a2658f3c6493f664125b0c5940db82a4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | We know the highest value would be at least $40$ but less than $50$ so we check $45$ , prime factorizing 45. We get $3^2 \cdot 5$ . We square this and get $81 \cdot 25$ . We know that $80 \cdot 25 = 2000$ , then we add 25 and get $2025$ , which does not satisfy our requirement of having the square less than $2023$ . The largest multiple of $5$ that satisfies this is $40$ and the smallest multiple of $5$ that works is $5$ so all multiples of $5$ from $5$ to $40$ satisfy the requirements. Now we divide each element of the set by $5$ and get $1-8$ so there are $\boxed{8}$ solutions. | A | 8 |
613300bbad713a1527d2802dbf19ad74 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | Let's use the triangle inequality. We know that for a triangle, the sum of the 2 shorter sides must always be longer than the longest side. This is because if the longest side were to be as long as the sum of the other sides, or longer, we would only have a line.
Similarly, for a convex quadrilateral, the sum of the shortest 3 sides must always be longer than the longest side. Thus, the answer is $\frac{26}{2}-1=13-1=\boxed{12}$ | D | 12 |
613300bbad713a1527d2802dbf19ad74 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | Say the chosen side is $a$ and the other sides are $b,c,d$
By the Generalised Polygon Inequality, $a<b+c+d$ . We also have $a+b+c+d=26\Rightarrow b+c+d=26-a$
Combining these two, we get $a<26-a\Rightarrow a<13$
The largest length that satisfies this is $a=\boxed{12}$ | D | 12 |
613300bbad713a1527d2802dbf19ad74 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | The quadrilateral can by cyclic only when it is an isosceles triangle. Without Loss of Generality, lets assume that this quadrilateral is a trapezoid. We can assume this as if we inscribe a trapezoid in a triangle, the base can be the diameter of the circle which is the longest chord in the circle, therefore maximizing the side length. By Brahmagupta's Formula, the area of the quadrilateral is defined by $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semi-perimeter. If the perimeter of the quadrilateral is $26$ , then the semi-perimeter will be $13$ . The area of the quadrilateral must be positive so the difference between the semi-perimeter and a side length must be greater than $0$ as otherwise, the area will be $0$ or negative. Therefore, the longest a side can be in this quadrilateral is $\boxed{12}$ | D | 12 |
613300bbad713a1527d2802dbf19ad74 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | This is an AMC 10 problem 4, so there is no need for any complex formulas. The largest singular side length from a quadrilateral comes from a trapezoid. So we can set the $2$ sides of the trapezoid equal to $4$ . Next we can split the trapezoid into $5$ triangles, where each base length of the triangle equals $4$ . So the top side equals $8$ , and the bottom side length equals $4+4+4$ $=$ $\boxed{12}$ ~ kabbybear | D | 12 |
a0794e532ec989d50607630576c1ba1c | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_5 | How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$
$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$ | Prime factorizing this gives us $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$
$10^{15}$ gives us $15$ digits and $243$ gives us $3$ digits. $15+3=\text{\boxed{18}$ | E | 18 |
a0794e532ec989d50607630576c1ba1c | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_5 | How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$
$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$ | Multiplying it out, we get that $8^5 \cdot 5^{10} \cdot 15^5 = 243000000000000000$ . Counting, we have the answer is $\text{\boxed{18}$ ~andliu766 | E | 18 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$ | Each of the vertices is counted $3$ times because each vertex is shared by three different edges.
Each of the edges is counted $2$ times because each edge is shared by two different faces.
Since the sum of the integers assigned to all vertices is $21$ , the final answer is $21\times3\times2=\boxed{126}$ | D | 126 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$ | Note that each vertex is counted $2\times 3=6$ times. Thus, the answer is $21\times6=\boxed{126}$ | D | 126 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$ | Just set one vertice equal to $21$ , it is trivial to see that there are $3$ faces with value $42$ , and $42 \cdot 3=\boxed{126}$ | D | 126 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$ | Since there are 8 vertices in a cube, there are $\dfrac{21}4$ vertices for two edges. There are $4$ edges per face, and $6$ faces in a cube, so the value of the cube is $\dfrac{21}4 \cdot 24 = \boxed{126}$ | D | 126 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$ | Set each vertex to value 1, so the sum of the vertices is 8. We find that the value of the cube, if all vertices are 1, is 48. We conclude that the value of the cube is 6 times the value of the sum of the vertices. Therefore, we choose $21\times6=\boxed{126}$ | D | 126 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$ | The wording of the problem implies that the answer should hold for any valid combination of integers. Thus, we choose the numbers $21, 0, 0, 0, 0, 0, 0, 0$ , which are indeed $8$ integers that add to $21$ . Doing this, we find three edges that have a value of $21$ , and from there, we get three faces with a value of $42$ (while the other three faces have a value of $0$ ). Adding the three faces together, we get $42+42+42 = \boxed{126}$ | D | 126 |
dcf55bfafb3d3e10c1e41b1657dc0ab0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. Bread is done when its internal temperature is $200$ degrees Fahrenheit. What is this in degrees on the Breadus scale?
$\textbf{(A) }33\qquad\textbf{(B) }34.5\qquad\textbf{(C) }36\qquad\textbf{(D) }37.5\qquad\textbf{(E) }39$ | To solve this question, you can use $y = mx + b$ where the $x$ is Fahrenheit and the $y$ is Breadus. We have $(110,0)$ and $(350,100)$ . We want to find the value of $y$ in $(200,y)$ that falls on this line. The slope for these two points is $\frac{5}{12}$ $y = \frac{5}{12}x + b$ . Solving for $b$ using $(110, 0)$ $\frac{550}{12} = -b$ . We get $b = \frac{-275}{6}$ . Plugging in $(200, y), \frac{1000}{12}-\frac{550}{12}=y$ . Simplifying, $\frac{450}{12} = \boxed{37.5}$ | D | 37.5 |
dcf55bfafb3d3e10c1e41b1657dc0ab0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. Bread is done when its internal temperature is $200$ degrees Fahrenheit. What is this in degrees on the Breadus scale?
$\textbf{(A) }33\qquad\textbf{(B) }34.5\qquad\textbf{(C) }36\qquad\textbf{(D) }37.5\qquad\textbf{(E) }39$ | Let $^\circ B$ denote degrees Breadus. We notice that $200^\circ F$ is $90^\circ F$ degrees to $0^\circ B$ , and $150^\circ F$ to $100^\circ B$ . This ratio is $90:150=3:5$ ; therefore, $200^\circ F$ will be $\dfrac3{3+5}=\dfrac38$ of the way from $0$ to $100$ , which is $\boxed{37.5.}$ | D | 37.5. |
dcf55bfafb3d3e10c1e41b1657dc0ab0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. Bread is done when its internal temperature is $200$ degrees Fahrenheit. What is this in degrees on the Breadus scale?
$\textbf{(A) }33\qquad\textbf{(B) }34.5\qquad\textbf{(C) }36\qquad\textbf{(D) }37.5\qquad\textbf{(E) }39$ | From $110$ to $350$ degrees Fahrenheit, the Breadus scale goes from $1$ to $100$ $110$ to $350$ degrees is a span of $240$ , and we can use this to determine how many Fahrenheit each Breadus unit is worth. $240$ divided by $100$ is $2.4$ , so each Breadus unit is $2.4$ Fahrenheit, starting at $110$ Fahrenheit. For example, $1$ degree on the Breadus scale is $110 + 2.4$ , or $112.4$ Fahrenheit. Using this information, we can figure out how many Breadus degrees $200$ Fahrenheit is. $200-110$ is $90$ , so we divide $90$ by $2.4$ to find the answer, which is $\boxed{37.5}$ | D | 37.5 |
dcf55bfafb3d3e10c1e41b1657dc0ab0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. Bread is done when its internal temperature is $200$ degrees Fahrenheit. What is this in degrees on the Breadus scale?
$\textbf{(A) }33\qquad\textbf{(B) }34.5\qquad\textbf{(C) }36\qquad\textbf{(D) }37.5\qquad\textbf{(E) }39$ | We note that the range of F temperatures that $0-100$ $\text{Br}^\circ$ represents is $350-110 = 240$ $\text{F}^\circ$ $200$ $\text{F}^\circ$ is $(200-110) = 90$ $\text{F}^\circ$ along the way to getting to $240$ $\text{F}^\circ$ , the end of this range, or $90/240 = 9/24 = 3/8 = 0.375$ of the way. Therefore if we switch to the Br scale, we are $0.375$ of the way to $100$ from $0$ , or at $\boxed{37.5}^\circ$ | D | 37.5 |
dcf55bfafb3d3e10c1e41b1657dc0ab0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. Bread is done when its internal temperature is $200$ degrees Fahrenheit. What is this in degrees on the Breadus scale?
$\textbf{(A) }33\qquad\textbf{(B) }34.5\qquad\textbf{(C) }36\qquad\textbf{(D) }37.5\qquad\textbf{(E) }39$ | We have the points $(0, 110)$ and $(100, 350)$ . We want to find $(x, 200)$ . The equation of the line is $y=\frac{12}{5}x+110$ . We use this to find $x=\frac{75}{2}=37.5$ , or $\boxed{37.5}$ .
~MC413551 | D | 37.5 |
77fb83402afdf5d51537d2f43b0ce047 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_9 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$ | Do careful casework by each month. In the month and the date, we need a $0$ , a $3$ , and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\boxed{9}$ .
For curious readers, the numbers (in chronological order) are: | E | 9 |
77fb83402afdf5d51537d2f43b0ce047 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_9 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$ | There is one $3$ , so we need one more (three more means that either the month or units digit of the day is $3$ ). For the same reason, we need one more $0$
If $3$ is the units digit of the month, then the $0$ can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two digits must match ( $11, 22$ ). For the second (tens digit of the day), we must have the other two be $1$ , as a month can't start with $2$ or $0$ . There are $3$ successes this way.
If $3$ is the tens digit of the day, then $0$ can be either the tens digit of the month or the units digit of the day. For the first case, $1$ must go in the other slots. For the second, the other two slots must be $1$ as well. There are $2$ successes here.
If $3$ is the units digit of the day, then $0$ could go in any of the $3$ remaining slots again. If it's the tens digit of the day, then the other digits must be $1$ . If $0$ is the units digit of the day, then the other two slots must both be $1$ . If $0$ is the tens digit of the month, then the other two slots can be either both $1$ or both $2$ . In total, there are $4$ successes here.
Summing through all cases, there are $3 + 2 + 4 = \boxed{9}$ dates. | E | 9 |
77fb83402afdf5d51537d2f43b0ce047 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_9 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$ | We start with $2023----$ we need an extra $0$ and an extra $3$ . So we have at least one of those extras in the days, except we can have the month $03$ . We now have $6$ possible months $01,02,03,10,11,12$ . For month $1$ we have two cases, we now have to add in another 1, and the possible days are $13,31$ . For month $2$ we need an extra $2$ so we can have the day $23$ note that we can't use $32$ because it is to large. Now for month $3$ we can have any number and multiply it by $11$ so we have the solution $11,22$ . For October we need a $1$ and a $3$ so we have $13,31$ as our choices. For November we have two choices which are $03,30$ .Now for December we have $0$ options. Summing $2+1+2+2+2$ we get $\boxed{9}$ solutions. | E | 9 |
95e0b26ee26a8b0a1e053bf6decdc7b0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$ | Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously.
We can write the following equations:
\[\frac{ax+11}{a+1}=x+1\qquad (1)\] \[\frac{ax+33}{a+3}=x+2\qquad (2)\]
Multiplying $(x+1)$ by $(a+1)$ and solving, we get: \[ax+11=ax+a+x+1\] \[11=a+x+1\] \[a+x=10\qquad (3)\]
Multiplying $(2)$ by $(a+3)$ and solving, we get: \[ax+33=ax+2a+3x+6\] \[33=2a+3x+6\] \[2a+3x=27\qquad (4)\]
Solving the system of equations for $(3)$ and $(4)$ , we find that $a=3$ and $x=\boxed{7}$ | D | 7 |
95e0b26ee26a8b0a1e053bf6decdc7b0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$ | Suppose Maureen took $n$ tests with an average of $m$
If she takes another test, her new average is $\frac{(nm+11)}{(n+1)}=m+1$
Cross-multiplying: $nm+11=nm+n+m+1$ , so $n+m=10$
If she takes $3$ more tests, her new average is $\frac{(nm+33)}{(n+3)}=m+2$
Cross-multiplying: $nm+33=nm+2n+3m+6$ , so $2n+3m=27$
But $2n+3m$ can also be written as $2(n+m)+m=20+m$ . Therefore $m=27-20=\boxed{7}$ | D | 7 |
95e0b26ee26a8b0a1e053bf6decdc7b0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$ | Let $s$ represent the sum of Maureen's test scores previously and $t$ be the number of scores taken previously.
So, $\frac{s+11}{t+1} = \frac{s}{t}+1$ and $\frac{s+33}{t+3} = \frac{s}{t}+2$
We can use the first equation to write $s$ in terms of $t$
We then substitute this into the second equation: $\frac{-t^2+10t+33}{t+3} = \frac{-t^2+10}{t}+2$
From here, we solve for t, getting $t=3$
We substitute this to get $s=21$
Therefore, the solution to the problem is $\frac{21}{3}=$ $\boxed{7}$ | D | 7 |
95e0b26ee26a8b0a1e053bf6decdc7b0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$ | Let's consider all the answer choices. If the average is $8$ , then, we can assume that all her test choices were $8$ . We can see that she must have gotten $8$ twice, in order for another score of $11$ to bring her average up by one. However, adding three $11$ 's will not bring her score up to 10. Continuing this process for the answer choices, we see that the answer is $\boxed{7}$ ~andliu766 | D | 7 |