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a46bf0f51976accb16fff061ecc83bd1 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_11 | Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle); [/asy]
$\text{(A)}\ 20 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30$ | The largest sum occurs when $13$ is placed in the center. This sum is $13 + 10 + 1 = 13 + 7 + 4 = \boxed{24}$ . Note: Two other common sums, $18$ and $21$ , are also possible. | D | 24 |
a46bf0f51976accb16fff061ecc83bd1 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_11 | Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle); [/asy]
$\text{(A)}\ 20 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30$ | Since the horizontal sum equals the vertical sum, twice this sum will be the sum
of the five numbers plus the number in the center. When the center number is $13$ , the sum is the largest, \[[10 + 4 + 1 + 7 + 2(13)]=2S\\ 48=2S\\ S=\boxed{24}\] The other
four numbers are divided into two pairs with equal sums. | D | 24 |
7210f5fc431832e4261f573a36cb5c8e | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_12 | The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$ . To the nearest whole percent, what percent of its games did the team lose?
$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$ | The ratio means that for every $11$ games won, $4$ are lost, so the team has won $11x$ games, lost $4x$ games, and played $15x$ games for some positive integer $x$ . The percentage of games lost is just $\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{27}$ | B | 27 |
7210f5fc431832e4261f573a36cb5c8e | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_12 | The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$ . To the nearest whole percent, what percent of its games did the team lose?
$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$ | The Won/Lost ratio is 11/4 so, for some number $N$ , the team won $11N$ games and lost $4N$ games. Thus, the team played $15N$ games and the fraction of games lost is $\dfrac{4N}{15N}=\dfrac{4}{15}\approx 0.27=\boxed{27}.$ -Clara Garza | null | 27 |
de34a6015ed7a4acd2ba9f66451ea740 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_13 | The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?
$\text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 29 \qquad \text{(E)}\ 30$ | First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals $(20)(15)+(15)(16)=540$ . The total amount of everyone's ages can be found from the average age, $17\cdot40=680$ . Then you do $680-540=140$ to find the sum of the adult's ages. The average age of an adult is divided among the five of them, $140\div5=\boxed{28}$ | C | 28 |
2cdac7351b3957e82b2a1726f7a4c29a | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_14 | In trapezoid $ABCD$ , the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is
[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8)); label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$16$",(8,0),S); label("$3$",(4,1.5),E); [/asy]
$\text{(A)}\ 27 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 34 \qquad \text{(E)}\ 48$ | [asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((12,3)--(12,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8)); label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$8$",(8,0),S); label("$3$",(4,1.5),E); label("$4$",(2,0),S); label("$4$",(14,0),S); label("$5$",(0,0)--(4,3),NW); label("$5$",(12,3)--(16,0),NE); [/asy]
There is a rectangle present, with both horizontal bases being $8$ units in length. The excess units on the bottom base must then be $16-8=8$ . The fact that $AB$ and $CD$ are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of $8$ units, so each is $4$ units. The triangle has a hypotenuse of $5$ , because the triangles are $3-4-5$ right triangles. So, the sides of the trapezoid are $8$ $5$ $16$ , and $5$ . Adding those up gives us the perimeter, $8 + 5 + 16 + 5 = 13 + 21 = \boxed{34}$ units. | D | 34 |
699f13be205dc2747b58726ff68b654e | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_15 | Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$ | There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \cdot 3 \cdot 4 = 60$ license plates.
Adding $2$ letters to the start gives $7\cdot 3 \cdot 4 = 84$ plates.
Adding $2$ letters to the middle gives $5 \cdot 5 \cdot 4 = 100$ plates.
Adding $2$ letters to the end gives $5 \cdot 3 \cdot 6 = 90$ plates.
Adding a letter to the start and middle gives $6 \cdot 4 \cdot 4 = 96$ plates.
Adding a letter to the start and end gives $6 \cdot 3 \cdot 5 = 90$ plates.
Adding a letter to the middle and end gives $5 \cdot 4 \cdot 5 = 100$ plates.
You can get at most $100$ license plates total, giving an additional $100 - 60 = 40$ plates, making the answer $\boxed{40}$ | D | 40 |
699f13be205dc2747b58726ff68b654e | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_15 | Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$ | Using the same logic as above, the number of combinations of plates is simply the product of the size of each set of letters.
In general, when three numbers have the same fixed sum, their product will be maximal when they are as close together as possible. This is a 3D analogue of the fact that a rectangle with fixed perimeter maximizes its area when the sides are equal (ie when it becomes a square). In this case, no matter where you add the letters, there will be $5 + 3 + 4 + 2 = 14$ letters in total. If you divide them as evenly as possible among the three groups, you get $5, 5, 4$ , which is a possible situation.
As before, the answer is $5 \cdot 5 \cdot 4 - 5 \cdot 3 \cdot 4 = 40$ , and the correct choice is $\boxed{40}$ | D | 40 |
699f13be205dc2747b58726ff68b654e | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_15 | Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$ | Before new letters were added, five different letters could have been chosen for the first position, three for the second, and four for the third. This means that $5\cdot 3\cdot 4=60$ plates could have been made.
If two letters are added to the second set, then $5\cdot 5\cdot 4=100$ plates can be made. If one letter is added to each of the second and third sets, then $5\cdot 4\cdot 5=100$ plates can be made. None of the other four ways to place the two letters will create as many plates. So, $100-60=\boxed{40}$ | A | 40 |
659a2b3dfeb170e3502fa44d5ad9db71 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_16 | Tori's mathematics test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered 70% of the arithmetic, 40% of the algebra, and 60% of the geometry problems correctly, she did not pass the test because she got less than 60% of the problems right. How many more problems would she have needed to answer correctly to earn a 60% passing grade?
$\text{(A)}\ 1 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 11$ | First, calculate how many of each type of problem she got right:
Arithmetic: $70\% \cdot 10 = 0.70 \cdot 10 = 7$
Algebra: $40\% \cdot 30 = 0.40 \cdot 30 = 12$
Geometry: $60\% \cdot 35 = 0.60 \cdot 35 = 21$
Altogether, Tori answered $7 + 12 + 21 = 40$ questions correct.
To get a $60\%$ on her test overall, she needed to get $60\% \cdot 75 = 0.60 \cdot 75 = 45$ questions right.
Therefore, she needed to answer $45 - 40 = 5$ more questions to pass, so the correct answer is $\boxed{5}$ | B | 5 |
174456a04e1d0467eee87d396dfc3d26 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_17 | At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: $1\frac{1}{2}$ cups of flour, $2$ eggs, $3$ tablespoons butter, $\frac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will make only full recipes, not partial recipes.
Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may be left over.)
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 15$ | If $108$ students eat $2$ cookies on average, there will need to be $108\cdot 2 = 216$ cookies. There are $15$ cookies per pan, meaning there needs to be $\frac{216}{15} = 14.4$ pans. However, since half-recipes are forbidden, we need to round up and make $\lceil \frac{216}{15}\rceil = 15$ pans.
$1$ pan requires $2$ eggs, so $15$ pans require $2\cdot 15 = 30$ eggs. Since there are $6$ eggs in a half dozen, we need $\frac{30}{6} = 5$ half-dozens of eggs, and the answer is $\boxed{5}$ | C | 5 |
0c007fa9bf834dab5df14f5c176f6993 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_18 | At Central Middle School the $108$ students who take the AMC8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of $15$ cookies, lists this items: $1\frac{1}{2}$ cups flour, $2$ eggs, $3$ tablespoons butter, $\frac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will make only full recipes, not partial recipes.
They learn that a big concert is scheduled for the same night and attendance will be down $25\%$ . How many recipes of cookies should they make for their smaller party?
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$ | If $108$ students eat $2$ cookies on average, there will need to be $108\cdot 2 = 216$ cookies. But with the smaller attendance, you will only need $100\% - 25\% = 75\%$ of these cookies, or $75\% \cdot 216 = 0.75\cdot 216 = 162$ cookies.
$162$ cookies requires $\frac{162}{15} = 10.8$ batches. However, since half-batches are forbidden, we must round up to get $\left\lceil \frac{162}{15} \right\rceil = 11$ batches, and the correct answer is $\boxed{11}$ | E | 11 |
0c007fa9bf834dab5df14f5c176f6993 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_18 | At Central Middle School the $108$ students who take the AMC8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of $15$ cookies, lists this items: $1\frac{1}{2}$ cups flour, $2$ eggs, $3$ tablespoons butter, $\frac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will make only full recipes, not partial recipes.
They learn that a big concert is scheduled for the same night and attendance will be down $25\%$ . How many recipes of cookies should they make for their smaller party?
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$ | If there were $108$ students before, with the $25\%$ of students missing, there will be $75\%$ of $108$ students left. This is $75\% \cdot 108 = 0.75 \cdot 108 = 81$ students. These students eat $81 \cdot 2 = 162$ cookies. Follow the logic of the second paragraph above to find that there needs to be $11$ batches, and the correct answer is $\boxed{11}$ | E | 11 |
e11460c7d660273e82e3f5d3ec8a72cd | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_19 | At Central Middle School, the 108 students who take the AMC 8 meet in the evening to talk about food and eat an average of two cookies apiece. Hansel and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists these items: $1\frac{1}{2}$ cups flour, $2$ eggs, $3$ tablespoons butter, $\frac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will make full recipes, not partial recipes.
Hansel and Gretel must make enough pans of cookies to supply 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be leftover, of course.)
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$ | For $216$ cookies, you need to make $\frac{216}{15} = 14.4$ pans. Since fractional pans are forbidden, round up to make $\lceil \frac{216}{15} \rceil = 15$ pans.
There are $3$ tablespoons of butter per pan, meaning $3 \cdot 15 = 45$ tablespoons of butter are required for $15$ pans.
Each stick of butter has $8$ tablespoons, so we need $\frac{45}{8} = 5.625$ sticks of butter. However, we must round up again because partial sticks of butter are forbidden! Thus, we need $\lceil \frac{45}{8} \rceil = 6$ sticks of butter, and the answer is $\boxed{6}$ | B | 6 |
7a2a427174a05c8e0f7ccdf1decfc7b4 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_21 | The degree measure of angle $A$ is
[asy] unitsize(12); draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle); draw(arc((1,-10),(1+19/sqrt(461),-10+10/sqrt(461)),(25/17,-155/17),CCW)); draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW)); draw(arc((900/83,-400/83),(900/83+19/sqrt(461),-400/83+10/sqrt(461)),(900/83 - 9/sqrt(97),-400/83 + 4/sqrt(97)),CCW)); label(rotate(30)*"$40^\circ$",(2,-8.9),ENE); label("$100^\circ$",(21/3,-2/3),SE); label("$110^\circ$",(900/83,-317/83),NNW); label("$A$",(0,0),NW); [/asy]
$\text{(A)}\ 20 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 45$ | Angle-chasing using the small triangles:
Use the line below and to the left of the $110^\circ$ angle to find that the rightmost angle in the small lower-left triangle is $180 - 110 = 70^\circ$
Then use the small lower-left triangle to find that the remaining angle in that triangle is $180 - 70 - 40 = 70^\circ$
Use congruent vertical angles to find that the lower angle in the smallest triangle containing $A$ is also $70^\circ$
Next, use line segment $AB$ to find that the other angle in the smallest triangle containing $A$ is $180 - 100 = 80^\circ$
The small triangle containing $A$ has a $70^\circ$ angle and an $80^\circ$ angle. The remaining angle must be $180 - 70 - 80 = \boxed{30}$ | B | 30 |
7a2a427174a05c8e0f7ccdf1decfc7b4 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_21 | The degree measure of angle $A$ is
[asy] unitsize(12); draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle); draw(arc((1,-10),(1+19/sqrt(461),-10+10/sqrt(461)),(25/17,-155/17),CCW)); draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW)); draw(arc((900/83,-400/83),(900/83+19/sqrt(461),-400/83+10/sqrt(461)),(900/83 - 9/sqrt(97),-400/83 + 4/sqrt(97)),CCW)); label(rotate(30)*"$40^\circ$",(2,-8.9),ENE); label("$100^\circ$",(21/3,-2/3),SE); label("$110^\circ$",(900/83,-317/83),NNW); label("$A$",(0,0),NW); [/asy]
$\text{(A)}\ 20 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 45$ | The third angle of the triangle containing the $100^\circ$ angle and the $40^\circ$ angle is $180^\circ - 100^\circ - 40^\circ = 40^\circ$ . It follows that $A$ is the third angle of the triangle consisting of the found $40^\circ$ angle and the given $110^\circ$ angle. Thus, $A$ is a $180^\circ - 110^\circ - 40^\circ = 30^\circ$ angle, and so the answer is $\boxed{30}$ | B | 30 |
e01c55a61fe64e3a704ffa77512cac11 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_23 | Square $ABCD$ has sides of length 3. Segments $CM$ and $CN$ divide the square's area into three equal parts. How long is segment $CM$
[asy] pair A,B,C,D,M,N; A = (0,0); B = (0,3); C = (3,3); D = (3,0); M = (0,1); N = (1,0); draw(A--B--C--D--cycle); draw(M--C--N); label("$A$",A,SW); label("$M$",M,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,SE); label("$N$",N,S); [/asy]
$\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}$ | Since the square has side length $3$ , the area of the entire square is $9$
The segments divide the square into 3 equal parts, so the area of each part is $9 \div 3 = 3$
Since $\triangle CBM$ has area $3$ and base $CB = 3$ , using the area formula for a triangle:
$A_{tri} = \frac{1}{2}bh$
$3 = \frac{1}{2}3h$
$h = 2$
Thus, height $BM = 2$
Since $\triangle CBM$ is a right triangle, $CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \boxed{13}$ | C | 13 |
e01c55a61fe64e3a704ffa77512cac11 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_23 | Square $ABCD$ has sides of length 3. Segments $CM$ and $CN$ divide the square's area into three equal parts. How long is segment $CM$
[asy] pair A,B,C,D,M,N; A = (0,0); B = (0,3); C = (3,3); D = (3,0); M = (0,1); N = (1,0); draw(A--B--C--D--cycle); draw(M--C--N); label("$A$",A,SW); label("$M$",M,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,SE); label("$N$",N,S); [/asy]
$\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}$ | Connect $AC$ $S_\triangle AMC=S_\triangle ANC$ . To satisfied the three area is equal, we have $2S_\triangle AMC=S_\triangle BMC$ $2S_\triangle ANC=S_\triangle DNC$ . Thus, $AM=AN=\frac{1}{2}BM=\frac{1}{2}AB=1$ $BM=2,BC=3,MC=\boxed{13}$ | C | 13 |
1394219570afe5b386d2254e081954ea | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_24 | When $1999^{2000}$ is divided by $5$ , the remainder is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | Note that the units digits of the powers of 9 have a pattern: $9^1 = {\bf 9}$ $9^2 = 8{\bf 1}$ $9^3 = 72{\bf 9}$ $9^4 = 656{\bf 1}$ , and so on. Since all natural numbers with the same last digit have the same remainder when divided by 5, the entire number doesn't matter, just the last digit. For even powers of $9$ , the number ends in a $1$ . Since the exponent is even, the final digit is $1$ . Note that all natural numbers that end in $1$ have a remainder of $1$ when divided by $5$ . So, our answer is $\boxed{1}$ | B | 1 |
1394219570afe5b386d2254e081954ea | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_24 | When $1999^{2000}$ is divided by $5$ , the remainder is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | Write $1999$ as $2000-1$ . We are taking $(2000-1)^{2000} \mod{10}$ . Using the binomial theorem, we see that ALL terms in this expansion are divisible by $2000$ except for the very last term, which is just $(-1)^{2000}$ . This is clear because the binomial expansion is just choosing how many $2000$ s and how many $-1$ s there are for each term. Using this, we can take the entire polynomial $\mod{10}$ , which leaves just $(-1)^{2000}=\boxed{1}$ | B | 1 |
1394219570afe5b386d2254e081954ea | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_24 | When $1999^{2000}$ is divided by $5$ , the remainder is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | As $1999 \equiv -1 \pmod{5}$ , we have $1999^{2000} \equiv (-1)^{2000} \equiv 1 \pmod{5}$ . Thus, the answer is $\boxed{1}$ | B | 1 |
1394219570afe5b386d2254e081954ea | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_24 | When $1999^{2000}$ is divided by $5$ , the remainder is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | A sum/product of any two natural numbers has the same remainder, when divided by $n \in \{\mathbb N}$ (Error compiling LaTeX. Unknown error_msg) , as the sum/product of their remainders. Thus, we can use the basic definition of an exponent and view the problem as $1999 \cdot 1999 \cdot \cdot \cdot 1999$
Using the fact stated in the first sentence, we see that the remainder of $1999$ , when divided by $5$ , is $4$ . The problem can now be written viewed as finding the remainder of $4^{2000}$ when it is divided by $5$ , which is already much simpler.
Now, toying with the simplified problem a little, see notice that powers of $4$ alternate from ending in $4$ and $6$ . We notice that even powers of $4$ always end in $6$ , and also that $2000$ is even. Thus $4^{2000}$ must end in $6$ , which, when divided by $5$ gives a remainder of $\boxed{1}$ | B | 1 |
c698d30b6ec9548b8dbc79aa9a891c85 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_25 | Points $B$ $D$ , and $J$ are midpoints of the sides of right triangle $ACG$ . Points $K$ $E$ $I$ are midpoints of the sides of triangle $JDG$ , etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$ , then the total area of the shaded triangles is nearest
[asy] draw((0,0)--(6,0)--(6,6)--cycle); draw((3,0)--(3,3)--(6,3)); draw((4.5,3)--(4.5,4.5)--(6,4.5)); draw((5.25,4.5)--(5.25,5.25)--(6,5.25)); fill((3,0)--(6,0)--(6,3)--cycle,black); fill((4.5,3)--(6,3)--(6,4.5)--cycle,black); fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black); label("$A$",(0,0),SW); label("$B$",(3,0),S); label("$C$",(6,0),SE); label("$D$",(6,3),E); label("$E$",(6,4.5),E); label("$F$",(6,5.25),E); label("$G$",(6,6),NE); label("$H$",(5.25,5.25),NW); label("$I$",(4.5,4.5),NW); label("$J$",(3,3),NW); label("$K$",(4.5,3),S); label("$L$",(5.25,4.5),S); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$ | Since $\triangle FGH$ is fairly small relative to the rest of the diagram, we can make an underestimate by using the current diagram. All triangles are right-isosceles triangles.
$CD = \frac {CG}{2} = 3, DE = \frac{CD}{2} = \frac{3}{2}, EF = \frac{DE}{2} = \frac{3}{4}$
$CB = CD = 3, DK = DE = \frac{3}{2}, EL = EF = \frac{3}{4}$
$[CBD] = \frac{1}{2}3^2 = \frac{9}{2}$
$[DKE] = \frac{1}{2}(\frac{3}{2})^2 = \frac{9}{8}$
$[ELF] = \frac{1}{2}(\frac{3}{4})^2 = \frac{9}{32}$
The sum of the shaded regions is $\frac{9}{2} + \frac{9}{8} + \frac{9}{32} = \frac{189}{32} \approx 5.9$
$5.9$ is an underestimate, as some portion (but not all) of $\triangle FGH$ will be shaded in future iterations.
If you shade all of $\triangle FGH$ , this will add an additional $\frac{9}{32}$ to the area, giving $\frac{198}{32} \approx 6.2$ , which is an overestimate.
Thus, $6 \rightarrow \boxed{6}$ is the only answer that is both over the underestimate and under the overestimate. | A | 6 |
c698d30b6ec9548b8dbc79aa9a891c85 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_25 | Points $B$ $D$ , and $J$ are midpoints of the sides of right triangle $ACG$ . Points $K$ $E$ $I$ are midpoints of the sides of triangle $JDG$ , etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$ , then the total area of the shaded triangles is nearest
[asy] draw((0,0)--(6,0)--(6,6)--cycle); draw((3,0)--(3,3)--(6,3)); draw((4.5,3)--(4.5,4.5)--(6,4.5)); draw((5.25,4.5)--(5.25,5.25)--(6,5.25)); fill((3,0)--(6,0)--(6,3)--cycle,black); fill((4.5,3)--(6,3)--(6,4.5)--cycle,black); fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black); label("$A$",(0,0),SW); label("$B$",(3,0),S); label("$C$",(6,0),SE); label("$D$",(6,3),E); label("$E$",(6,4.5),E); label("$F$",(6,5.25),E); label("$G$",(6,6),NE); label("$H$",(5.25,5.25),NW); label("$I$",(4.5,4.5),NW); label("$J$",(3,3),NW); label("$K$",(4.5,3),S); label("$L$",(5.25,4.5),S); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$ | In iteration $1$ , congruent triangles $\triangle ABJ, \triangle BDJ,$ and $\triangle BDC$ are created, with one of them being shaded.
In iteration $2$ , three more congruent triangles are created, with one of them being shaded.
As the process continues indefnitely, in each row, $\frac{1}{3}$ of each triplet of new congruent triangles will be shaded. The "fourth triangle" at the top ( $\triangle FGH$ in the diagram) will gradually shrink,
leaving about $\frac{1}{3}$ of the area shaded. This means $\frac{1}{3}\left(\frac{1}{2}6\cdot 6\right) = 6$ square units will be shaded when the process goes on indefinitely, giving $\boxed{6}$ | A | 6 |
c698d30b6ec9548b8dbc79aa9a891c85 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_25 | Points $B$ $D$ , and $J$ are midpoints of the sides of right triangle $ACG$ . Points $K$ $E$ $I$ are midpoints of the sides of triangle $JDG$ , etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$ , then the total area of the shaded triangles is nearest
[asy] draw((0,0)--(6,0)--(6,6)--cycle); draw((3,0)--(3,3)--(6,3)); draw((4.5,3)--(4.5,4.5)--(6,4.5)); draw((5.25,4.5)--(5.25,5.25)--(6,5.25)); fill((3,0)--(6,0)--(6,3)--cycle,black); fill((4.5,3)--(6,3)--(6,4.5)--cycle,black); fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black); label("$A$",(0,0),SW); label("$B$",(3,0),S); label("$C$",(6,0),SE); label("$D$",(6,3),E); label("$E$",(6,4.5),E); label("$F$",(6,5.25),E); label("$G$",(6,6),NE); label("$H$",(5.25,5.25),NW); label("$I$",(4.5,4.5),NW); label("$J$",(3,3),NW); label("$K$",(4.5,3),S); label("$L$",(5.25,4.5),S); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$ | Using Solution 1 as a template, note that the sum of the areas forms a geometric series
$\frac{9}{2} + \frac{9}{8} + \frac{9}{32} + \frac{9}{128} + ...$
This is the sum of a geometric series with first term $a_1 = \frac{9}{2}$ and common ratio $r = \frac{1}{4}$ This is the easiest way to do this problem.
The sum of an infinite geometric series with $|r|<1$ is shown by the formula. $S_{\infty} = \frac{a_1}{1 - r}$ Insert the values to get $\frac{\frac{9}{2}}{1 - \frac{1}{4}} = \frac{9}{2}\cdot\frac{4}{3} = 6$ , giving an answer of $\boxed{6}$ | A | 6 |
c698d30b6ec9548b8dbc79aa9a891c85 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_25 | Points $B$ $D$ , and $J$ are midpoints of the sides of right triangle $ACG$ . Points $K$ $E$ $I$ are midpoints of the sides of triangle $JDG$ , etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$ , then the total area of the shaded triangles is nearest
[asy] draw((0,0)--(6,0)--(6,6)--cycle); draw((3,0)--(3,3)--(6,3)); draw((4.5,3)--(4.5,4.5)--(6,4.5)); draw((5.25,4.5)--(5.25,5.25)--(6,5.25)); fill((3,0)--(6,0)--(6,3)--cycle,black); fill((4.5,3)--(6,3)--(6,4.5)--cycle,black); fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black); label("$A$",(0,0),SW); label("$B$",(3,0),S); label("$C$",(6,0),SE); label("$D$",(6,3),E); label("$E$",(6,4.5),E); label("$F$",(6,5.25),E); label("$G$",(6,6),NE); label("$H$",(5.25,5.25),NW); label("$I$",(4.5,4.5),NW); label("$J$",(3,3),NW); label("$K$",(4.5,3),S); label("$L$",(5.25,4.5),S); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$ | Find the area of the bottom triangle, which is 4.5. Notice that the area of the triangles is divided by 4 every time. 4.5*5/4≈5.7, and 5.7+(1/4)≈5.9. We can clearly see that the sum is approaching answer choice $\boxed{6}$ | A | 6 |
c698d30b6ec9548b8dbc79aa9a891c85 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_25 | Points $B$ $D$ , and $J$ are midpoints of the sides of right triangle $ACG$ . Points $K$ $E$ $I$ are midpoints of the sides of triangle $JDG$ , etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$ , then the total area of the shaded triangles is nearest
[asy] draw((0,0)--(6,0)--(6,6)--cycle); draw((3,0)--(3,3)--(6,3)); draw((4.5,3)--(4.5,4.5)--(6,4.5)); draw((5.25,4.5)--(5.25,5.25)--(6,5.25)); fill((3,0)--(6,0)--(6,3)--cycle,black); fill((4.5,3)--(6,3)--(6,4.5)--cycle,black); fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black); label("$A$",(0,0),SW); label("$B$",(3,0),S); label("$C$",(6,0),SE); label("$D$",(6,3),E); label("$E$",(6,4.5),E); label("$F$",(6,5.25),E); label("$G$",(6,6),NE); label("$H$",(5.25,5.25),NW); label("$I$",(4.5,4.5),NW); label("$J$",(3,3),NW); label("$K$",(4.5,3),S); label("$L$",(5.25,4.5),S); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$ | Obviously, the area of $DBC$ is $\frac{1}{3}$ of $AJDC$ , the area of $DEA$ is $\frac{1}{3}$ of $JIED$ , if the progress is going to infinity, the shaded triangles will be $\frac{1}{3}$ of the triangle $ACG$ . However, 100 times is much enough. The answer is $\frac{1}{3}\times 6\times 6=\boxed{6}$ | A | 6 |
c4f227afce145e5df7246b1a455e4cba | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_2 | If $\begin{tabular}{r|l}a&b \\ \hline c&d\end{tabular} = \text{a}\cdot \text{d} - \text{b}\cdot \text{c}$ , what is the value of $\begin{tabular}{r|l}3&4 \\ \hline 1&2\end{tabular}$
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | Plugging in values for $a$ $b$ $c$ , and $d$ , we get
$a=3$ $b=4$ $c=1$ $d=2$
$a\times d=3\times2=6$
$b\times c=4\times1=4$
$6-4=2$
$\boxed{2}$ | E | 2 |
f3a0f2534d620901749ff46882d65153 | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_4 | How many triangles are in this figure? (Some triangles may overlap other triangles.)
[asy] draw((0,0)--(42,0)--(14,21)--cycle); draw((14,21)--(18,0)--(30,9)); [/asy]
$\text{(A)}\ 9 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 5$ | By inspection, we have that there $5$ triangles: Each of the $3$ small triangles, $1$ medium triangle made of the rightmost two small triangles, and the $1$ large triangle.
$\boxed{5}$ | E | 5 |
b66d7d95e38adbf47294ecdf69f79f2b | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_6 | Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is
[asy] for(int a=0; a<4; ++a) { for(int b=0; b<4; ++b) { dot((a,b)); } } draw((0,0)--(0,2)--(1,2)--(2,3)--(2,2)--(3,2)--(3,0)--(2,0)--(2,1)--(1,0)--cycle); [/asy]
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$ | We could count the area contributed by each square on the $3 \times 3$ grid:
Top-left: $0$
Top: Triangle with area $\frac{1}{2}$
Top-right: $0$
Left: Square with area $1$
Center: Square with area $1$
Right: Square with area $1$
Bottom-left: Square with area $1$
Bottom: Triangle with area $\frac{1}{2}$
Bottom-right: Square with area $1$
Adding all of these together, we get $6$ which is the same as $\boxed{6}$ | B | 6 |
b66d7d95e38adbf47294ecdf69f79f2b | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_6 | Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is
[asy] for(int a=0; a<4; ++a) { for(int b=0; b<4; ++b) { dot((a,b)); } } draw((0,0)--(0,2)--(1,2)--(2,3)--(2,2)--(3,2)--(3,0)--(2,0)--(2,1)--(1,0)--cycle); [/asy]
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$ | By Pick's Theorem, we get the formula, $A=I+\frac{b}{2}-1$ where $I$ is the number of lattice points in the interior and $b$ being the number of lattice points on the boundary. In this problem, we can see that $I=1$ and $B=12$ . Substituting gives us $A=1+\frac{12}{2}-1=6$ Thus, the answer is $\boxed{6}$ | B | 6 |
b66d7d95e38adbf47294ecdf69f79f2b | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_6 | Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is
[asy] for(int a=0; a<4; ++a) { for(int b=0; b<4; ++b) { dot((a,b)); } } draw((0,0)--(0,2)--(1,2)--(2,3)--(2,2)--(3,2)--(3,0)--(2,0)--(2,1)--(1,0)--cycle); [/asy]
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$ | Notice that the extra triangle on the top with area $1$ can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area $1$ . This creates a $2*3$ rectangle, with a area of $6$ . The answer is $\boxed{6}$ ~sakshamsethi | B | 6 |
54922f0a670e832b2a8f30c5f5d8b7d7 | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_8 | A child's wading pool contains 200 gallons of water. If water evaporates at the rate of 0.5 gallons per day and no other water is added or removed, how many gallons of water will be in the pool after 30 days?
$\text{(A)}\ 140 \qquad \text{(B)}\ 170 \qquad \text{(C)}\ 185 \qquad \text{(D)}\ 198.5 \qquad \text{(E)}\ 199.85$ | $30$ days multiplied by $0.5$ gallons a day results in $15$ gallons of water loss.
The remaining water is $200-15=185=\boxed{185}$ | C | 185 |
c3ac768fd749a34eb3bcc9fc3c1fd5dd | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_9 | For a sale, a store owner reduces the price of a $$10$ scarf by $20\%$ . Later the price is lowered again, this time by one-half the reduced price. The price is now
$\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qquad \text{(E)}\ 6.40\text{ dollars}$ | $100\%-20\%=80\%$
$10\times80\%=10\times0.8$
$10\times0.8=8$
$\frac{8}{2}=4=\boxed{4.00}$ | C | 4.00 |
c3ac768fd749a34eb3bcc9fc3c1fd5dd | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_9 | For a sale, a store owner reduces the price of a $$10$ scarf by $20\%$ . Later the price is lowered again, this time by one-half the reduced price. The price is now
$\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qquad \text{(E)}\ 6.40\text{ dollars}$ | The first discount has percentage 20, which is then discounted again for half of the already discounted price.
$100-20=80$
$\frac{80}{2}=40$
$40\%\times10=10\times0.4=4=\boxed{4.00}$ | C | 4.00 |
26d96b3888db74b27572655329e082b7 | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_10 | Each of the letters $\text{W}$ $\text{X}$ $\text{Y}$ , and $\text{Z}$ represents a different integer in the set $\{ 1,2,3,4\}$ , but not necessarily in that order. If $\dfrac{\text{W}}{\text{X}} - \dfrac{\text{Y}}{\text{Z}}=1$ , then the sum of $\text{W}$ and $\text{Y}$ is
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$ | There are different ways to approach this problem, and I'll start with the different factor of the numbers of the set $\{ 1,2,3,4\}$
$1$ has factor $1$
$2$ has factors $1$ and $2$
$3$ has factors $1$ and $3$
$4$ has factors $1$ $2$ , and $4$
From here, we note that even though all numbers have the factor $1$ , only $4$ has another factor other than $1$ in the set (ie. $2$
We could therefore have one fraction be $\frac{4}{2}$ and another $\frac{3}{1}$
The sum of the numerators is $4+3=7=\boxed{7}$ | E | 7 |
8691deed9afab388a6cb6058d4d5cc10 | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_11 | Harry has 3 sisters and 5 brothers. His sister Harriet has $\text{S}$ sisters and $\text{B}$ brothers. What is the product of $\text{S}$ and $\text{B}$
$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18$ | Harry has 3 sisters and 5 brothers. His sister, being a girl, would have 1 less sister and 1 more brother.
$S = 3-1=2$
$B = 5+1=6$
$S\cdot B = 2\times6=12=\boxed{12}$ | C | 12 |
194bcd117f9c40e5023276e24021a67c | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_12 | $2\left(1-\dfrac{1}{2}\right) + 3\left(1-\dfrac{1}{3}\right) + 4\left(1-\dfrac{1}{4}\right) + \cdots + 10\left(1-\dfrac{1}{10}\right)=$
$\text{(A)}\ 45 \qquad \text{(B)}\ 49 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$ | Taking the first product, we have
$\left(1-\frac{1}{2}\right)=\frac{1}{2}$
$\frac{1}{2}\times2=1$
Looking at the second, we get
$\left(1-\frac{1}{3}\right)=\frac{2}{3}$
$\frac{2}{3}\times3=2$
We seem to be going up by $1$
Just to check,
$1-\frac{1}{n}=\frac{n-1}{n}$
$\frac{n-1}{n}\times n=n-1$
Now that we have discovered the pattern, we have to find the last term.
$1-\frac{1}{10}=\frac{9}{10}$
$\frac{9}{10}\times10=9$
The sum of all numbers from $1$ to $9$ is
$\frac{9\cdot10}{2}=45=\boxed{45}$ | A | 45 |
70a2ea7ce2eabab923fb59e98aba27ac | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_14 | An Annville Junior High School, $30\%$ of the students in the Math Club are in the Science Club, and $80\%$ of the students in the Science Club are in the Math Club. There are 15 students in the Science Club. How many students are in the Math Club?
$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 40$ | If $80\%$ of the people in the science club of $15$ people are in the Math Club, $\frac{4}{5}\times15=12$ people are in the both the Math Club and the Science Club.
These $12$ people make up $30\%$ of the Math Club.
Setting up a proportion:
$12\cdot 1.00 =0.30\cdot x$
$\frac{12}{0.3} = 40=x$
There are $40=\boxed{40}$ people in the Math Club. | E | 40 |
23cbe7f60d2b2eefad8384b7fea228e6 | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_16 | Estimate the year in which the population of Nisos will be approximately 6,000.
$\text{(A)}\ 2050 \qquad \text{(B)}\ 2075 \qquad \text{(C)}\ 2100 \qquad \text{(D)}\ 2125 \qquad \text{(E)}\ 2150$ | We could triple the population every $25$ years and make a chart:
Year: 2000
Population: 200
Year: 2025
Population: 600
Year: 2050
Population: 1800
Year: 2075
Population: 5400
Year: 2100
Population: 16200
The closest year is 2075, or $\boxed{2075}$ | B | 2075 |
23cbe7f60d2b2eefad8384b7fea228e6 | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_16 | Estimate the year in which the population of Nisos will be approximately 6,000.
$\text{(A)}\ 2050 \qquad \text{(B)}\ 2075 \qquad \text{(C)}\ 2100 \qquad \text{(D)}\ 2125 \qquad \text{(E)}\ 2150$ | We could find out how many periods of 25 years we need to triple by dividing our total from our present number.
$\frac{6000}{200}=30$
The power of $3$ that $30$ is closest to is $27=3^3$
Therefore, after $3$ periods, we will be closest to $6000$
$\boxed{2075}$ | B | 2075 |
84e0d60108090ecae293df652333055c | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_17 | In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?
$\text{(A)}\ 50\text{ yrs.} \qquad \text{(B)}\ 75\text{ yrs.} \qquad \text{(C)}\ 100\text{ yrs.} \qquad \text{(D)}\ 125\text{ yrs.} \qquad \text{(E)}\ 150\text{ yrs.}$ | We can divide the total area by how much will be occupied per person:
$\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}$ can stay on the island at its maximum capacity.
We can divide 16600 by the current population in $1998$ which is 200 to see by what factor the population increases:
$\frac{16600}{200}=83$ -fold increase in population.
Thus, the population increases by a factor $83$ . This is very close to $3 \times 3 \times 3 \times 3 = 81$ , and so there are about $4$ triplings of the island's population.
It takes $4\times25=100=\boxed{100}$ years to triple the island's population four times in succession. | C | 100 |
84e0d60108090ecae293df652333055c | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_17 | In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?
$\text{(A)}\ 50\text{ yrs.} \qquad \text{(B)}\ 75\text{ yrs.} \qquad \text{(C)}\ 100\text{ yrs.} \qquad \text{(D)}\ 125\text{ yrs.} \qquad \text{(E)}\ 150\text{ yrs.}$ | We can continue the pattern, and because the pattern increases numbers rapidly, it won't be hard. $\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}$ can live on the island at its maximum capacity.
$1998: 200 \text{ people}$ $2023: 600 \text{ people}$ $2048: 1800 \text{ people}$ $2073: 5400 \text{ people}$ $2098: 16200 \text{ people}$ After the year $2098$ , it will not be possible for the next increase to occur, because when $16200$ is tripled, it is way more than the maximum capacity. Thus, the answer is $2098-1998 = 100$ , or $\boxed{100}$ | C | 100 |
a132045eff8d92f1681eecd8b312f43c | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_21 | $4\times 4\times 4$ cubical box contains 64 identical small cubes that exactly fill the box. How many of these small cubes touch a side or the bottom of the box?
$\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$ | Each small cube would have dimensions $1\times 1\times 1$ making each cube a unit cube.
If there are $16$ cubes per face and there are $5$ faces we are counting, we have $16\times 5= 80$ cubes.
Some cubes are on account of overlap between different faces.
We could reduce this number by subtracting the overlap areas, which could mean subtracting 4 cubes from each side and 12 from the bottom.
$80-(4\times 4+12)=80-28=52=\boxed{52}$ | B | 52 |
a132045eff8d92f1681eecd8b312f43c | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_21 | $4\times 4\times 4$ cubical box contains 64 identical small cubes that exactly fill the box. How many of these small cubes touch a side or the bottom of the box?
$\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$ | You can imagine removing the cubes that do not fit the description of the problem, forming a "square cup".
There are $4$ cubes in the center of the top face that do not fit the description. Remove those.
Once you remove those cubes on top, you must go down one level and remove the cubes in the same position on the second layer. Thus, $4$ more cubes are removed.
Finally, you repeat this on the third layer, for $4$ more cubes.
Once you do the top three layers, you will be on the bottom layer, and you don't remove any more cubes.
This means you removed $4 + 4 + 4 = 12$ cubes of the $4 \times 4 \times 4 = 64$ cubes.
Thus, $64 - 12 = 52$ cubes remain, and the answer is $\boxed{52}$ | B | 52 |
a132045eff8d92f1681eecd8b312f43c | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_21 | $4\times 4\times 4$ cubical box contains 64 identical small cubes that exactly fill the box. How many of these small cubes touch a side or the bottom of the box?
$\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$ | We can use casework to solve this.
Case $1$ : The cubes on the sides
There are four sides of a cube which means $64$ cubes, but there will be an overlap of $16$ cubes ( $4$ sides $\times$ $4$ cubes on each side).
So there are $16 \times 4-16=64-16=48$ cubes on the sides.
Case $2$ : The cubes on the bottom
There are $16$ cubes on the bottom, but by counting the cubes on the sides, $12$ of the cubes are already accounted for. $16-12=4$ cubes on the bottom that aren't already counted.
$48+4=\boxed{52}$ cubes touching at least one side. Since there is no top, we don't need to count the top face cubes. | B | 52 |
2a629927d0bbc98de8375047555ca2f8 | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_22 | Terri produces a sequence of positive integers by following three rules. She starts with a positive integer, then applies the appropriate rule to the result, and continues in this fashion.
Rule 1: If the integer is less than 10, multiply it by 9.
Rule 2: If the integer is even and greater than 9, divide it by 2.
Rule 3: If the integer is odd and greater than 9, subtract 5 from it.
A sample sequence: $23, 18, 9, 81, 76, \ldots .$
Find the $98^\text{th}$ term of the sequence that begins $98, 49, \ldots .$
$\text{(A)}\ 6 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 27 \qquad \text{(E)}\ 54$ | We could start by looking for a pattern.
$98, 49, 44, 22, 11, 6, 54, 27, 22, 11, 6, \ldots .$
From here, we see that we have a pattern of $22, 11, 6, 54, 27, \ldots .$ after $98, 49, 44$
Our problem is now really
Find the $95^\text{th}$ term of the sequence that goes $22, 11, 6, 54, 27, 22, 11, 6, 54, 27, 22, \ldots .$
There are 5 terms in each repetition of the pattern, and $95\equiv0\pmod{5}$ , so the answer is $\boxed{27}$ | D | 27 |
6338ecb2ee0ad7a18845b57b432fb4ad | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_24 | A rectangular board of 8 columns has squares numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result?
[asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy]
$\text{(A)}\ 36\qquad\text{(B)}\ 64\qquad\text{(C)}\ 78\qquad\text{(D)}\ 91\qquad\text{(E)}\ 120$ | The numbers that are shaded are the triangular numbers, which are numbers in the form $\frac{(n)(n+1)}{2}$ for positive integers. They can also be generated by starting with $1$ , and adding $1, 2, 3, 4...$ as in the description of the problem.
Squares that have the same remainder after being divided by $8$ will be in the same column. Thus, we want to find when the last remainder, from $0$ to $7$ , is found.
So, instead of adding $1, 2, 3, 4, 5, 6, 7, 8, 9, 10...$ , we can effectively either add $1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3...$ or subtract $7, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5...$ if we are only concerned about remainders when divided by $8$ . We will pick the number that keeps the terms on the list between $1$ and $8$ . We get:
$1$
$1 + 2 = 3$
$3 + 3 = 6$
$6 - 4 = 2$
$2 + 5 = 7$
$7 - 2 = 5$
$5 - 1 = 4$
$4 + 0 = 4$
$4 + 1 = 5$
$5 + 2 = 7$
$7 - 5 = 2$
$2 + 4 = 6$
$6 - 3 = 3$
$3 - 2 = 1$
$1 - 1 = 0$
Finally, a term with $0$ is found, and checking, all numbers $1$ through $7$ are also on the right side of the list. This means the last term in our sequence is the first time that column $8$ is shaded. There are $15$ terms in the sequence, leading to an answer of $\frac{15\cdot 16}{2} = 120$ , which is choice $\boxed{120}$ | E | 120 |
6338ecb2ee0ad7a18845b57b432fb4ad | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_24 | A rectangular board of 8 columns has squares numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result?
[asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy]
$\text{(A)}\ 36\qquad\text{(B)}\ 64\qquad\text{(C)}\ 78\qquad\text{(D)}\ 91\qquad\text{(E)}\ 120$ | Note that the triangular numbers up to $120$ are $1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120$ . When you divide each of those numbers by $8$ , all remainders must be present. We first search for number(s) that are evenly divisible by $8$ ; if two such numbers exist, we search for numbers that leave a remainder of $1$ , etc.
Quickly scanning the list, only $6, 10, 28, 36, 66, 78$ and $120$ are even. That smaller list doesn't have any multiples of $8$ until it hits $120$ . So $\boxed{120,}$ must be the answer. | E | 120, |
6338ecb2ee0ad7a18845b57b432fb4ad | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_24 | A rectangular board of 8 columns has squares numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result?
[asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy]
$\text{(A)}\ 36\qquad\text{(B)}\ 64\qquad\text{(C)}\ 78\qquad\text{(D)}\ 91\qquad\text{(E)}\ 120$ | The numbers shaded are triangular numbers of the form $\frac{(n)(n+1)}{2}$ . For this number to be divisible by $8$ , the numerator must be divisible by $16$ . Since only one of $n$ and $n+1$ can be even, only one of them can have factors of $2$ . Therefore, the first time the whole expression is divisible by $8$ is when either $n+1=16$ or when $n=16$ . This gives $n=15$ as the first time $\frac{n(n+1)}{2}$ is divisible by $8$ , which gives $120$ . No other triangular number lower than that is divisible by $8$ , and thus the $8^{th}$ column on the checkerboard won't be filled until then. That gives $\boxed{120}$ as the right answer. | E | 120 |
b0b1607d5299cac951057e77d42ab09e | https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_25 | Three generous friends, each with some money, redistribute the money as followed:
Amy gives enough money to Jan and Toy to double each amount has.
Jan then gives enough to Amy and Toy to double their amounts.
Finally, Toy gives enough to Amy and Jan to double their amounts.
If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 216\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 288$ | If Toy had $36$ dollars at the beginning, then after Amy doubles his money, he has $36 \times 2 = 72$ dollars after the first step.
Then Jan doubles his money, and Toy has $72 \times 2 = 144$ dollars after the second step.
Then Toy doubles whatever Amy and Jan have. Since Toy ended up with $36$ , he spent $144 - 36 = 108$ to double their money. Therefore, just before this third step, Amy and Jan must have had $108$ dollars in total. And, just before this step, Toy had $144$ dollars. Altogether, the three had $144 + 108 = 252$ dollars, and the correct answer is $\boxed{252}$ | D | 252 |
abc11ac04ebff49b73b7d4ea3914fd3e | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_1 | $\dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000} =$
$\text{(A)}\ 0.0026 \qquad \text{(B)}\ 0.0197 \qquad \text{(C)}\ 0.1997 \qquad \text{(D)}\ 0.26 \qquad \text{(E)}\ 1.997$ | Convert each fraction to a decimal.
$\dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000}$
$0.1+0.09+0.009+0.0007=0.1997$
$\boxed{0.1997}$ | C | 0.1997 |
a8122b47f8b709b67afd92c7708e314b | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_2 | Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. What is the largest number Ahn can get?
$\text{(A)}\ 200 \qquad \text{(B)}\ 202 \qquad \text{(C)}\ 220 \qquad \text{(D)}\ 380 \qquad \text{(E)}\ 398$ | The smallest two-digit integer he can subtract from $200$ is $10$ . This will give the largest result for that first operation, and doubling it will keep it as the largest number possible.
\[200-10=190\] \[190\times2=380\]
$\boxed{380}$ | D | 380 |
5526892e6a6a036f766b8d85638c45c9 | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_3 | Which of the following numbers is the largest?
$\text{(A)}\ 0.97 \qquad \text{(B)}\ 0.979 \qquad \text{(C)}\ 0.9709 \qquad \text{(D)}\ 0.907 \qquad \text{(E)}\ 0.9089$ | Using the process of elimination:
Tenths digit:
All tenths digits are equal, at $9$
Hundreths digit: $A$ $B$ , and $C$ all have the same hundreths digit of $7$ , and it is greater than the hundredths of $D$ or $E$ (which is $0$ ). Eliminate both $D$ and $E$
Thousandths digit: $B$ has the largest thousandths digit of the remaining answers, and is the correct answer. $A$ has an "invisible" thousandths digit of $0$ , while $C$ also has a thousdandths digit of $0$
$\boxed{0.979}$ | B | 0.979 |
686650a28bcdfaf5b35b8598f2c4be54 | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_4 | Julie is preparing a speech for her class. Her speech must last between one-half hour and three-quarters of an hour. The ideal rate of speech is 150 words per minute. If Julie speaks at the ideal rate, which of the following number of words would be an appropriate length for her speech?
$\text{(A)}\ 2250 \qquad \text{(B)}\ 3000 \qquad \text{(C)}\ 4200 \qquad \text{(D)}\ 4350 \qquad \text{(E)}\ 5650$ | Since there are $60$ minutes in an hour, we need a number of words between $(\frac{1}{2}\times 60)\times150=4500$ and $(\frac{3}{4}\times60)\times150=6750$ . The only such choice is $\boxed{5650}$ | E | 5650 |
a50a17419115f93336089cc195078d7f | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_5 | There are many two-digit multiples of 7, but only two of the multiples have a digit sum of 10. The sum of these two multiples of 7 is
$\text{(A)}\ 119 \qquad \text{(B)}\ 126 \qquad \text{(C)}\ 140 \qquad \text{(D)}\ 175 \qquad \text{(E)}\ 189$ | Writing out all two digit numbers that have a digital sum of $10$ , you get $19, 28, 37, 46, 55, 64, 73, 82,$ and $91$ . The two numbers on that list that are divisible by $7$ are $28$ and $91$ . Their sum is $28+91=119$ , choice $\boxed{119}$ | A | 119 |
a50a17419115f93336089cc195078d7f | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_5 | There are many two-digit multiples of 7, but only two of the multiples have a digit sum of 10. The sum of these two multiples of 7 is
$\text{(A)}\ 119 \qquad \text{(B)}\ 126 \qquad \text{(C)}\ 140 \qquad \text{(D)}\ 175 \qquad \text{(E)}\ 189$ | Writing out all the two digit multiples of $7$ , you get $14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,$ and $91$ . Again you find $28$ and $91$ have a digital sum of $10$ , giving answer $\boxed{119}$ | A | 119 |
688c2fac5f3995b00ed6be42ae6efd7e | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_6 | In the number $74982.1035$ the value of the place occupied by the digit 9 is how many times as great as the value of the place occupied by the digit 3?
$\text{(A)}\ 1,000 \qquad \text{(B)}\ 10,000 \qquad \text{(C)}\ 100,000 \qquad \text{(D)}\ 1,000,000 \qquad \text{(E)}\ 10,000,000$ | The digit $9$ is $5$ places to the left of the digit $3$ . Thus, it has a place value that is $10^5 = 100,000$ times greater. $\boxed{100,000}$ | C | 100,000 |
688c2fac5f3995b00ed6be42ae6efd7e | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_6 | In the number $74982.1035$ the value of the place occupied by the digit 9 is how many times as great as the value of the place occupied by the digit 3?
$\text{(A)}\ 1,000 \qquad \text{(B)}\ 10,000 \qquad \text{(C)}\ 100,000 \qquad \text{(D)}\ 1,000,000 \qquad \text{(E)}\ 10,000,000$ | The digit $9$ is in the $100$ s place. The digit $3$ is in the $\frac{1}{1000}$ ths place. The ratio of these two numbers is $\frac{100}{\frac{1}{1000}} = 100\times1000 = 100,000$ , and the answer is $\boxed{100,000}$ | C | 100,000 |
806412d432830921882917a5df6ac3ba | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_7 | The area of the smallest square that will contain a circle of radius 4 is
$\text{(A)}\ 8 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$ | Draw a square circumscribed around the circle. (Alternately, the circle is inscribed in the square.) If the circle has radius $4$ , it has diameter $8$ . Two of the diameters of the circle will run parallel to the sides of the square. Thus, the smallest square that contains it has side length $8$ , and area $8\times8=64$ $\boxed{64}$ | D | 64 |
36ffca03dc7af7928c9d87a73c70b434 | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_8 | Walter gets up at 6:30 a.m., catches the school bus at 7:30 a.m., has 6 classes that last 50 minutes each, has 30 minutes for lunch, and has 2 hours additional time at school. He takes the bus home and arrives at 4:00 p.m. How many minutes has he spent on the bus?
$\text{(A)}\ 30 \qquad \text{(B)}\ 60 \qquad \text{(C)}\ 75 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 120$ | There are $4\frac{1}{2}$ hours from 7:30 a.m. to noon, and $4$ hours from noon to 4 p.m., meaning Walter was away from home for a total of $8\frac{1}{2}$ hours, or $8.5\times 60 = 510$ minutes.
Tallying up the times he was not on the bus, he has $6\times 50 = 300$ minutes in classes, $30$ minutes at lunch, and $2\times 60 = 120$ minutes of "additional time". That is a total of $300 + 30 + 120 = 450$ minutes that Walter was away from home, but not on the bus.
Therefore, Walter spent $510 - 450 = 60$ minutes on the bus, and the answer is $\boxed{60}$ | B | 60 |
55c7e5b4adadf6a5a8cc3bccf716b698 | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_12 | $\angle 1 + \angle 2 = 180^\circ$
$\angle 3 = \angle 4$
Find $\angle 4.$
[asy] pair H,I,J,K,L; H = (0,0); I = 10*dir(70); J = I + 10*dir(290); K = J + 5*dir(110); L = J + 5*dir(0); draw(H--I--J--cycle); draw(K--L--J); draw(arc((0,0),dir(70),(1,0),CW)); label("$70^\circ$",dir(35),NE); draw(arc(I,I+dir(250),I+dir(290),CCW)); label("$40^\circ$",I+1.25*dir(270),S); label("$1$",J+0.25*dir(162.5),NW); label("$2$",J+0.25*dir(17.5),NE); label("$3$",L+dir(162.5),WNW); label("$4$",K+dir(-52.5),SE); [/asy] $\text{(A)}\ 20^\circ \qquad \text{(B)}\ 25^\circ \qquad \text{(C)}\ 30^\circ \qquad \text{(D)}\ 35^\circ \qquad \text{(E)}\ 40^\circ$ | Using the left triangle, we have:
$\angle 1 + 70 + 40 = 180$
$\angle 1 = 180 - 110$
$\angle 1 = 70$
Using the given fact that $\angle 1 + \angle 2 = 180$ , we have $\angle 2 = 180 - 70 = 110$
Finally, using the right triangle, and the fact that $\angle 3 = \angle 4$ , we have:
$\angle 2 + \angle 3 + \angle 4 = 180$
$110 + \angle 3 + \angle 4 = 180$
$110 + \angle 4+ \angle 4 = 180$
$2\angle 4 = 70$
$\angle 4 = 35$
Thus, the answer is $\boxed{35}$ | D | 35 |
066a84a8325a8865ea1d236ac81c0771 | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_13 | Three bags of jelly beans contain 26, 28, and 30 beans. The ratios of yellow beans to all beans in each of these bags are $50\%$ $25\%$ , and $20\%$ , respectively. All three bags of candy are dumped into one bowl. Which of the following is closest to the ratio of yellow jelly beans to all beans in the bowl?
$\text{(A)}\ 31\% \qquad \text{(B)}\ 32\% \qquad \text{(C)}\ 33\% \qquad \text{(D)}\ 35\% \qquad \text{(E)}\ 95\%$ | In bag $A$ , there are $26$ jellybeans, and $50\%$ are yellow. That means there are $26\times 50\% = 26\times 0.50 = 13$ yellow jelly beans in this bag.
In bag $B$ , there are $28$ jellybeans, and $25\%$ are yellow. That means there are $28\times 25\% = 28\times 0.25 = 7$ yellow jelly beans in this bag.
In bag $C$ , there are $30$ jellybeans, and $20\%$ are yellow. That means there are $30\times 20\% = 30\times 0.20 = 6$ yellow jelly beans in this bag.
In all three bags, there are $13 + 7 + 6 = 26$ yellow jelly beans in total , and $26 + 28 + 30 = 84$ jelly beans of all types in total.
Thus, $\frac{26}{84} = \frac{26}{84}\cdot 100\% = \frac{13 }{ 42} \cdot 100\% = 30.9\%$ of all jellybeans are yellow. Thus, the correct answer is $\boxed{31}$ | A | 31 |
fd650ece4cf52e342ef1d6fdd585ff85 | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_14 | There is a set of five positive integers whose average (mean) is 5, whose median is 5, and whose only mode is 8. What is the difference between the largest and smallest integers in the set?
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | When these numbers are ordered in ascending order, 5, the median, falls right in the middle, which is the third integer from the left. Since there is a unique mode of 8, both integers to the right of 5 must be 8s. Since the mean is 5, the sum of the integers is 25, which means the 2 leftmost integers have to sum to 4. 2 and 2 does not work because that would result in two modes. However, 1 and 3 does, and so our answer is 8-1=7. $\boxed{7}$ | D | 7 |
25ae69dd2501a1005c50778ac8ac988b | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_17 | A cube has eight vertices (corners) and twelve edges. A segment, such as $x$ , which joins two vertices not joined by an edge is called a diagonal. Segment $y$ is also a diagonal. How many diagonals does a cube have?
[asy] draw((0,3)--(0,0)--(3,0)--(5.5,1)--(5.5,4)--(3,3)--(0,3)--(2.5,4)--(5.5,4)); draw((3,0)--(3,3)); draw((0,0)--(2.5,1)--(5.5,1)--(0,3)--(5.5,4),dashed); draw((2.5,4)--(2.5,1),dashed); label("$x$",(2.75,3.5),NNE); label("$y$",(4.125,1.5),NNE); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$ | On each face, there are $2$ diagonals like $x$ . There are $6$ faces on a cube. Thus, there are $2\times 6 = 12$ diagonals that are "x-like".
Every "y-like" diagonal must connect the bottom of the cube to the top of the cube. Thus, for each of the $4$ bottom vertices of the cube, there is a different "y-like" diagonal. So there are $4$ "y-like" diagonals.
This gives a total of $12 + 4 = 16$ diagonals on the cube, which is answer $\boxed{16}$ | E | 16 |
25ae69dd2501a1005c50778ac8ac988b | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_17 | A cube has eight vertices (corners) and twelve edges. A segment, such as $x$ , which joins two vertices not joined by an edge is called a diagonal. Segment $y$ is also a diagonal. How many diagonals does a cube have?
[asy] draw((0,3)--(0,0)--(3,0)--(5.5,1)--(5.5,4)--(3,3)--(0,3)--(2.5,4)--(5.5,4)); draw((3,0)--(3,3)); draw((0,0)--(2.5,1)--(5.5,1)--(0,3)--(5.5,4),dashed); draw((2.5,4)--(2.5,1),dashed); label("$x$",(2.75,3.5),NNE); label("$y$",(4.125,1.5),NNE); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$ | There are $8$ vertices on a cube(A, B, C, D, E, F, G, H). If you pick one vertice, it will have 7 line segments associated with it, so you will have $\frac{8\cdot7}{2} = 28$ segments within the cube. The division by $2$ is necessary because you counted both the segment from $A$ to $B$ and the segment from $B$ to $A$
But not all $28$ of these segments are diagonals. Some are edges. There are $4$ edges on the top, $4$ edges on the bottom, and $4$ edges that connect the top to the bottom. So there are $12$ edges total, meaning that there are $28 - 12 = 16$ segments that are not edges. All of these segments are diagonals, and thus the answer is $\boxed{16}$ | E | 16 |
25ae69dd2501a1005c50778ac8ac988b | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_17 | A cube has eight vertices (corners) and twelve edges. A segment, such as $x$ , which joins two vertices not joined by an edge is called a diagonal. Segment $y$ is also a diagonal. How many diagonals does a cube have?
[asy] draw((0,3)--(0,0)--(3,0)--(5.5,1)--(5.5,4)--(3,3)--(0,3)--(2.5,4)--(5.5,4)); draw((3,0)--(3,3)); draw((0,0)--(2.5,1)--(5.5,1)--(0,3)--(5.5,4),dashed); draw((2.5,4)--(2.5,1),dashed); label("$x$",(2.75,3.5),NNE); label("$y$",(4.125,1.5),NNE); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$ | Consider picking one point on the corner of the cube. That point has $3$ "x-like" diagonals , and $1$ "y-like" diagonal that ends on the opposite vertex. Thus, each vertex has $3 + 1 = 4$ diagonals associated with it. There are $8$ vertices on the cube, giving a total of $8 \cdot 4 = 32$ diagonals.
However, each diagonal was counted as both a "starting point" and an "ending point". So there are really $\frac{32}{2} = 16$ diagonals, giving an answer of $\boxed{16}$ | E | 16 |
4baf8de49c866adeba93eed91e96ee8e | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_18 | At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for $$5$ . This week they are on sale at 5 boxes for $$4$ . The percent decrease in the price per box during the sale was closest to
$\text{(A)}\ 30\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 65\%$ | Last week, each box was $\frac{5}{4} = 1.25$
This week, each box is $\frac{4}{5} = 0.80$
Percent decrease is given by $\frac{X_{old} - X_{new}}{X_{old}} \cdot 100\%$
This, the percent decrease is $\frac{1.25 - 0.8}{1.25}\cdot 100\% = \frac{45}{125} \cdot 100\% = 36\%$ , which is closest to $\boxed{35}$ | B | 35 |
d1d65a6f2d0f5b31db64dceb76fada5a | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_19 | If the product $\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9$ , what is the sum of $a$ and $b$
$\text{(A)}\ 11 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 37$ | Notice that the numerator of the first fraction cancels out the denominator of the second fraction, and the numerator of the second fraction cancels out the denominator of the third fraction, and so on.
The only numbers left will be $a$ in the numerator from the last fraction and $2$ in the denominator from the first fraction. (The $b$ will cancel with the numerator of the preceeding number.) Thus, $\frac{a}{2} = 9$ , and $a=18$
Since the numerator is always one more than the denominator, $b = a-1 = 17$ , and $a+ b = 18 + 17 = 35$ , giving an answer of $\boxed{35}$ | D | 35 |
d1d65a6f2d0f5b31db64dceb76fada5a | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_19 | If the product $\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9$ , what is the sum of $a$ and $b$
$\text{(A)}\ 11 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 37$ | Find a pattern. If $a=4$ , then the expression is just the first two terms, which is $2$
If $a=5$ , then the expression is the first three terms, giving $2.5$
If $a=6$ , the expression is the first four terms, giving $3$
If $a=7$ , the expression will be the first five terms, giving $3.5$
Conjecture that the expression is always going to equal $\frac{a}{2}$ , and thus when $a=18$ , the expression will be $9$ , as desired.
As above, when $a=18$ $b=17$ , and the sum is $35$ , or $\boxed{35}$ | D | 35 |
ede47c1ddf7642fefe543038fe1fe99f | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_21 | Each corner cube is removed from this $3\text{ cm}\times 3\text{ cm}\times 3\text{ cm}$ cube. The surface area of the remaining figure is
[asy] draw((2.7,3.99)--(0,3)--(0,0)); draw((3.7,3.99)--(1,3)--(1,0)); draw((4.7,3.99)--(2,3)--(2,0)); draw((5.7,3.99)--(3,3)--(3,0)); draw((0,0)--(3,0)--(5.7,0.99)); draw((0,1)--(3,1)--(5.7,1.99)); draw((0,2)--(3,2)--(5.7,2.99)); draw((0,3)--(3,3)--(5.7,3.99)); draw((0,3)--(3,3)--(3,0)); draw((0.9,3.33)--(3.9,3.33)--(3.9,0.33)); draw((1.8,3.66)--(4.8,3.66)--(4.8,0.66)); draw((2.7,3.99)--(5.7,3.99)--(5.7,0.99)); [/asy]
$\text{(A)}\ 19\text{ sq.cm} \qquad \text{(B)}\ 24\text{ sq.cm} \qquad \text{(C)}\ 30\text{ sq.cm} \qquad \text{(D)}\ 54\text{ sq.cm} \qquad \text{(E)}\ 72\text{ sq.cm}$ | The original cube has $6$ square surfaces that each have an area of $3^2 = 9$ , for a toal surface area of $6\cdot 9 = 54$
Since no two corner cubes touch, we can examine the effect of removing each corner cube individually.
Each corner cube contribues $3$ faces each of surface area $1$ to the big cube, so the surface area is decreased by $3$ when the cube is removed.
However, when the cube is removed, $3$ faces on the 3x3x3 cube will be revealed, increasing the surface area by $3$
Thus, the surface area does not change with the removal of a corner cube, and it remains $54$ , which is answer $\boxed{54}$ | D | 54 |
c0ac4ea6aabc85f60ea4f5a8df38d94c | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_22 | A two-inch cube $(2\times 2\times 2)$ of silver weighs 3 pounds and is worth 200 dollars. How much is a three-inch cube of silver worth?
$\textbf{(A) }\text{300 dollars} \qquad \textbf{(B) }\text{375 dollars} \qquad \textbf{(C) }\text{450 dollars} \qquad \textbf{(D) }\text{560 dollars}\qquad \textbf{(E) }\text{675 dollars}$ | The two-inch cube has a volume of $8$ cubic inches, and the three-inch cube has a volume of $27$ cubic inches. Thus, the three-inch cube has a weight that is $\frac{27}{8}$ times that of the two-inch cube. Then its value is $\frac{27}{8} \cdot 200 = \boxed{675}$ | E | 675 |
50dccf35fdd83fc0e1ff884034688605 | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_23 | There are positive integers that have these properties:
The product of the digits of the largest integer with both properties is
$\text{(A)}\ 7 \qquad \text{(B)}\ 25 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 60$ | Five-digit numbers will have a minimum of $1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$ as the sum of their squares if the five digits are distinct and non-zero. If there is a zero, it will be forced to the left by rule #2.
No digit will be greater than $7$ , as $8^2 = 64$
Trying four digit numbers $WXYZ$ , we have $w^2 + x^2 + y^2 + z^2 = 50$ with $0 < w < x < y < z < 8$
$z=7$ will not work, since the other digits must be at least $1^2 + 2^2 + 3^2 = 14$ , and the sum of the squares would be over $50$
$z=6$ will give $w^2 + x^2 + y^2 = 14$ $(w,x,y) = (1,2,3)$ will work, giving the number $1236$ . No other number with $z=6$ will work, as $w, x,$ and $y$ would have to be greater.
$z=5$ will give $w^2 + x^2 + y^2 = 25$ $y=4$ forces $x=3$ and $w=0$ , which has a leading zero, and then we have 345 which is a 3-digit number.
$z=4$ can only give the number $1234$ , which does not satisfy the condition of the problem.
Thus, the number in question is $1236$ , and the product of the digits is $36$ , giving $\boxed{36}$ as the answer. | C | 36 |
f01f2be9dc7283ab1e45f0a022b0b0da | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_25 | All of the even numbers from 2 to 98 inclusive, excluding those ending in 0, are multiplied together. What is the rightmost digit (the units digit) of the product?
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | All the tens digits of the product will be irrelevant to finding the units digit. Thus, we are searching for the units digit of $(2\cdot 4\cdot 6 \cdot 8) \cdot (2 \cdot 4 \cdot 6 \cdot 8) \cdot (2\cdot 4\cdot 6 \cdot 8) \cdot ...$
There will be $10$ groups of $4$ numbers. The number now can be rewritten as $(2\cdot 4 \cdot 6 \cdot 8)^{10}$
Multiplying, we get $384^{10}$
Again, we can disregard the tens and hundreds digit of $384$ , since we only want the units digit of the number, leaving $4^{10}$
Now, we try to find a pattern to the units digit of $4^n$ . To compute this quickly, we once again discard all tens digits and higher.
$4^1 = 4$
$4^2 = 4\cdot 4 = 1\underline{6}$ , discard the $1$
$4^3 = 1 \cdot 4 = \underline{4}$
$4^4 = 4 \cdot 4 = 1\underline{6}$ , discard the $1$
$4^5 = 1 \cdot 4 = \underline{4}$
Those equalities are, in reality, congruences $\mod {10}$
Thus, the pattern of the units digits is $\{4, 6, 4, 6, 4, 6, 4, 6, 4, 6\}$ . The cycle repeats so that term $n$ is the same as term $n+2$ . The tenth number in the cycle is $6$ , giving an answer of $\boxed{6}$ | D | 6 |
f01f2be9dc7283ab1e45f0a022b0b0da | https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_25 | All of the even numbers from 2 to 98 inclusive, excluding those ending in 0, are multiplied together. What is the rightmost digit (the units digit) of the product?
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | Again, the value we seek is equal to $(2\cdot 4\cdot 6\cdot 8)^{10}\mod 10$ .
We can use equivalence to simplify.
$(2\cdot 4\cdot 6\cdot 8)^{10}\mod 10$
$\equiv (2\cdot 4\cdot (-4)\cdot (-2))^{10} \mod 10$
$\equiv 64^{10} \mod 10$
$\equiv 4^{10} \mod 10$
$\equiv 2^{20} \mod 10$
$\equiv 8^6\cdot 2^2 \mod 10$
$\equiv (-2)^6\cdot 2^2 \mod 10$
$\equiv 2^8 \mod 10$
$\equiv 256 \mod 10$
$\equiv 6 \mod 10$
Thus our answer is $\boxed{6}$ | D | 6 |
63fec9f1e4abb24e1e7c327741f3b2ac | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_1 | How many positive factors of 36 are also multiples of 4?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$ | The factors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18,$ and $36$
The multiples of $4$ up to $36$ are $4, 8, 12, 16, 20, 24, 28, 32$ and $36$
Only $4, 12$ and $36$ appear on both lists, so the answer is $3$ , which is option $\boxed{3}$ | B | 3 |
63fec9f1e4abb24e1e7c327741f3b2ac | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_1 | How many positive factors of 36 are also multiples of 4?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$ | $36 = 4^1 \cdot 3^2$ . All possible factors of $36$ will be here, except for ones divisible by $2$ and not by $4$ $(1+1)\cdot (2+1) = 6$ . Subtract factors not divisible by $4$ , which are $1$ $3^1$ , and $3^2$ $6-3=3$ , which is $\boxed{3}$ | B | 3 |
63fec9f1e4abb24e1e7c327741f3b2ac | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_1 | How many positive factors of 36 are also multiples of 4?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$ | Divide $36$ by $4$ , and the remaining factors, when multiplied by $4$ , will be factors of $36$
$36 \div 4 = 9$ , which has $3$ factors, giving us option $\boxed{3}$ | B | 3 |
b0624b7dcc596925ec5b3b9390c24cbd | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_3 | The 64 whole numbers from 1 through 64 are written, one per square, on a checkerboard (an 8 by 8 array of 64 squares). The first 8 numbers are written in order across the first row, the next 8 across the second row, and so on. After all 64 numbers are written, the sum of the numbers in the four corners will be
$\text{(A)}\ 130 \qquad \text{(B)}\ 131 \qquad \text{(C)}\ 132 \qquad \text{(D)}\ 133 \qquad \text{(E)}\ 134$ | Obviously $1$ is in the top left corner, $8$ is in the top right corner, and $64$ is in the bottom right corner. To find the bottom left corner, subtract $7$ from $64$ which is $57$ . Adding the results gives $1+8+57+64=130$ which is answer $\boxed{130}$ | A | 130 |
5c7a9827b345a135765615ca81dcf18f | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_6 | What is the smallest result that can be obtained from the following process?
$\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 56$ | Since we want the smallest possible result, and we are only adding and multiplying positive numbers over $1$ , we can "prune" the set to the three smallest numbers $\{3,5,7\}$ . Using bigger numbers will create bigger sums and bigger products.
From there, compute the $3$ ways you can do the two operations:
$(3+5)7 = 8\cdot 7 = 56$
$(3 + 7)5 = 10\cdot 5 = 50$
$(7+5)3 = 12\cdot 3 = 36$
The smallest number is 36, giving an answer of $\boxed{36}$ | C | 36 |
f56c8cb33daf457a01a3b34806eedb78 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_7 | Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month. If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, then in how many months from that time will they have the same number of goldfish?
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | Call this month "Month 0". Make a table of the fish that Brent and Gretel have each month.
$\text{Month / Brent / Gretel}$
$\text{0 / 4 / 128}$
$\text{1 / 16 / 256}$
$\text{2 / 64 / 512}$
$\text{3 / 256 / 1024}$
$\text{4 / 1024 / 2048}$
$\text{5 / 4096 / 4096}$
You could create a similar table without doing all of the calculations by converting all the goldfish into powers of $2$ . In this table, you could increase Brent's goldfish by two powers of $2$ , while increasing Gretel's fish by one power of $2$
Either way, in $5$ months they will have the same number of fish, giving an answer of $\boxed{5}$ | B | 5 |
f56c8cb33daf457a01a3b34806eedb78 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_7 | Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month. If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, then in how many months from that time will they have the same number of goldfish?
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | Create two equations.
Brent starts with $4$ goldfish and they quadruple each month, giving us $y=4\cdot 4^{x}$ or $y=2^{2x+2}$
Gretel starts with $128$ goldfish and they double each month, giving us $y=128 \cdot 2^{x}$ or $y=2^{x+7}$
Using substitution, $2^{2x+2}=2^{x+7}$ . Thus $2x+2=x+7$ giving us $x=5$ , which is $\boxed{5}$ | B | 5 |
92ae407cdd0856ba32f6cadbe6f3ec19 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_8 | Points $A$ and $B$ are 10 units apart. Points $B$ and $C$ are 4 units apart. Points $C$ and $D$ are 3 units apart. If $A$ and $D$ are as close as possible, then the number of units between them is
$\text{(A)}\ 0 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 11 \qquad \text{(E)}\ 17$ | If $AB = 10$ and $BC=4$ , then $(10 - 4) \le AC \le (10 + 4)$ by the triangle inequality . In the triangle inequality, the equality is only reached when the "triangle" $ABC$ is really a degenerate triangle, and $ABC$ are collinear.
Simplifying, this means the smallest value $AC$ can be is $6$
Applying the triangle inequality on $ACD$ with $AC = 6$ and $CD = 3$ , we know that $6 - 3 \le AD \le 6 + 3$ when $AC$ is minimized. If $AC$ were larger, then $AD$ could be larger, but we want the smallest $AD$ possible, and not the largest. Thus, $AD$ must be at least $3$ , but cannot be smaller than $3$ . Therefore, $\boxed{3}$ is the answer. | B | 3 |
276471e91846e052402a420ce4eed499 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_9 | If 5 times a number is 2, then 100 times the reciprocal of the number is
$\text{(A)}\ 2.5 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 250 \qquad \text{(E)}\ 500$ | If $5$ times a number is $2$ , then $5x = 2 \rightarrow x = \frac{2}{5}$ , and the number is $\frac{2}{5}$
If the number is $\frac{2}{5}$ , then $100$ times its reciprocal is $100$ times $\frac{5}{2}$ , which is $250$ , giving an answer of $\boxed{250}$ | D | 250 |
78a12b1f7d10fa443ca1c83d5a9fc2fd | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_10 | When Walter drove up to the gasoline pump, he noticed that his gasoline tank was 1/8 full. He purchased 7.5 gallons of gasoline for 10 dollars. With this additional gasoline, his gasoline tank was then 5/8 full. The number of gallons of gasoline his tank holds when it is full is $\text{(A)}\ 8.75 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11.5 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 22.5$ | The tank started at $\frac{1}{8}$ full, and ended at $\frac{5}{8}$ full. Therefore, Walter filled $\frac{5}{8} - \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$ of the tank.
If Walter fills half the tank with $7.5$ gallons, then Walter can fill two halves of the tank (or a whole tank) with $7.5 \times 2 = 15$ gallons, giving an answer of $\boxed{15}$ | D | 15 |
57f9f0d767255796b1fae80ae249d65d | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_11 | Let $x$ be the number \[0.\underbrace{0000...0000}_{1996\text{ zeros}}1,\] where there are 1996 zeros after the decimal point. Which of the following expressions represents the largest number?
$\text{(A)}\ 3+x \qquad \text{(B)}\ 3-x \qquad \text{(C)}\ 3\cdot x \qquad \text{(D)}\ 3/x \qquad \text{(E)}\ x/3$ | Estimate each of the options.
$A$ will give a number that is just over $3$
$B$ will give a number that is just under $3$ . This eliminates $B$ , because $A$ is bigger.
$C$ will give a number that is barely over $0$ , since it is three times a tiny number. This eliminates $C$ , because $A$ is bigger.
$D$ will give a huge number. $\frac{1}{x}$ will get very, very large in magnitude when $x$ gets close to zero. You can see this by examining the sequence $x=0.1$ , which gives $10$ as the reciprocal, $x=0.01$ , which gives $100$ as the reciprocal, and $x=0.001$ , which gives $1000$ as the reciprocal. Thus, $D$ will be huge, and this eliminates $A$
$E$ will give a small number, since you're dividing a tiny number into thirds. This eliminates $E$ , and gives $\boxed{3}$ as the answer. | D | 3 |
4e76af7cb120e1e69a4116ae1a5f1422 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_12 | What number should be removed from the list \[1,2,3,4,5,6,7,8,9,10,11\] so that the average of the remaining numbers is $6.1$
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | Adding all of the numbers gives us $\frac{11\cdot12}{2}=66$ as the current total. Since there are $11$ numbers, the current average is $\frac{66}{11}=6$ . We need to take away a number from the total and then divide the result by $10$ because there will only be $10$ numbers left to give an average of $6.1$ . Setting up the equation:
$\frac{66-x}{10}=6.1$
$66 - x = 61$
$x = 5$
Thus, the answer is $\boxed{5}$ | B | 5 |
4e76af7cb120e1e69a4116ae1a5f1422 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_12 | What number should be removed from the list \[1,2,3,4,5,6,7,8,9,10,11\] so that the average of the remaining numbers is $6.1$
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | Similar to the first solution, the current total is $66$ . Since there are $11$ numbers on the list, taking $1$ number away will leave $10$ numbers. If those $10$ numbers have an average of $6.1$ , then those $10$ numbers must have a sum of $10 \times 6.1 = 61$ . Thus, the number that was removed must be $66 - 61 = 5$ , and the answer is $\boxed{5}$ | B | 5 |
72d0a161d19a19a6c1ba7e1460d0b774 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_13 | In the fall of 1996, a total of 800 students participated in an annual school clean-up day. The organizers of the event expect that in each of the years 1997, 1998, and 1999, participation will increase by 50% over the previous year. The number of participants the organizers will expect in the fall of 1999 is
$\text{(A)}\ 1200 \qquad \text{(B)}\ 1500 \qquad \text{(C)}\ 2000 \qquad \text{(D)}\ 2400 \qquad \text{(E)}\ 2700$ | If the participation increases by $50\%$ , then it is the same as saying participation is multipled by a factor of $100\% + 50\% = 1 + 0.5 = 1.5$
In 1997, participation will be $800 \cdot 1.5 = 1200$
In 1998, participation will be $1200 \cdot 1.5 = 1800$
In 1999, participation will be $1800 \cdot 1.5 = 2700$ , giving an answer of $\boxed{2700}$ | E | 2700 |
72d0a161d19a19a6c1ba7e1460d0b774 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_13 | In the fall of 1996, a total of 800 students participated in an annual school clean-up day. The organizers of the event expect that in each of the years 1997, 1998, and 1999, participation will increase by 50% over the previous year. The number of participants the organizers will expect in the fall of 1999 is
$\text{(A)}\ 1200 \qquad \text{(B)}\ 1500 \qquad \text{(C)}\ 2000 \qquad \text{(D)}\ 2400 \qquad \text{(E)}\ 2700$ | Since the percentage increase is the same each year, this is an example of exponential growth with a base of $1.5$ . In three years, there will be $1.5^3 = \frac{27}{8}$ times as many participants. Multiplying this by the $800$ current participants, there are $2700$ participants, and the answer is $\boxed{2700}$ | E | 2700 |
b3970a6ad9aa6295b5d7caad9f4b3c8d | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_14 | Six different digits from the set \[\{ 1,2,3,4,5,6,7,8,9\}\] are placed in the squares in the figure shown so that the sum of the entries in the vertical column is 23 and the sum of the entries in the horizontal row is 12.
The sum of the six digits used is
[asy] unitsize(18); draw((0,0)--(1,0)--(1,1)--(4,1)--(4,2)--(1,2)--(1,3)--(0,3)--cycle); draw((0,1)--(1,1)--(1,2)--(0,2)); draw((2,1)--(2,2)); draw((3,1)--(3,2)); label("$23$",(0.5,0),S); label("$12$",(4,1.5),E); [/asy]
$\text{(A)}\ 27 \qquad \text{(B)}\ 29 \qquad \text{(C)}\ 31 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 35$ | Looking at the vertical column, the three numbers sum to $23$ . If we make the numbers on either end $9$ and $8$ in some order, the middle number will be $6$ . This is the minimum for the intersection.
Looking at the horizontal row, the four numbers sum to $12$ . If we minimize the three numbers on the right to $123$ , the first number has a maximum value of $6$ . This is the maximum for the intersection
Thus, the minimum of the intersection is $6$ , and the maximum of the intersection is $6$ . This means the intersection must be $6$ , and the other numbers must be $9$ and $8$ in the column, and $123$ in the row. The sum of all the numbers is $12 + 23 - 6 = 29$ , and the answer is $\boxed{29}$ | B | 29 |
98910278f29f6a820328ff5cad3039a6 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_15 | The remainder when the product $1492\cdot 1776\cdot 1812\cdot 1996$ is divided by 5 is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | To determine a remainder when a number is divided by $5$ , you only need to look at the last digit. If the last digit is $0$ or $5$ , the remainder is $0$ . If the last digit is $1$ or $6$ , the remainder is $1$ , and so on.
To determine the last digit of $1492\cdot 1776\cdot 1812\cdot 1996$ , you only need to look at the last digit of each number in the product. Thus, we compute $2\cdot 6\cdot 2\cdot 6 = 12^2 = 144$ . The last digit of the number $1492\cdot 1776\cdot 1812\cdot 1996$ is also $4$ , and thus the remainder when the number is divided by $5$ is also $4$ , which gives an answer of $\boxed{4}$ | E | 4 |
98910278f29f6a820328ff5cad3039a6 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_15 | The remainder when the product $1492\cdot 1776\cdot 1812\cdot 1996$ is divided by 5 is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | $1492\cdot 1776\cdot 1812\cdot 1996 \mod 5$
$\equiv 2\cdot 1\cdot 2\cdot 1 \mod 5$
$=4$ , which is option $\boxed{4}$ | E | 4 |
259881424aafc718f47a2db47b2848ab | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_16 | $1-2-3+4+5-6-7+8+9-10-11+\cdots + 1992+1993-1994-1995+1996=$
$\text{(A)}\ -998 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 998$ | Put the numbers in groups of $4$
$(1-2-3+4)+(5-6-7+8)+(9-10-11+ 12) + \cdots + (1993-1994-1995+1996)$
The first group has a sum of $0$
The second group increases the two positive numbers on the end by $1$ , and decreases the two negative numbers in the middle by $1$ . Thus, the second group also has a sum of $0$
Continuing the pattern, every group has a sum of $0$ , and thus the entire sum is $0$ , giving an answer of $\boxed{0}$ | C | 0 |
259881424aafc718f47a2db47b2848ab | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_16 | $1-2-3+4+5-6-7+8+9-10-11+\cdots + 1992+1993-1994-1995+1996=$
$\text{(A)}\ -998 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 998$ | Let any term of the series be $t_n$ . Realize that at every $n\equiv0 \pmod4$ , the sum of the series is 0. For $t_{1996}$ we know $1996\equiv0 \pmod4$ so the solution is $\boxed{0}$ | C | 0 |
5a4ffa69eae668df9a09040f609fe9ab | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_17 | Figure $OPQR$ is a square. Point $O$ is the origin, and point $Q$ has coordinates (2,2). What are the coordinates for $T$ so that the area of triangle $PQT$ equals the area of square $OPQR$
[asy] pair O,P,Q,R,T; O = (0,0); P = (2,0); Q = (2,2); R = (0,2); T = (-4,0); draw((-5,0)--(3,0)); draw((0,-1)--(0,3)); draw(P--Q--R); draw((-0.2,-0.8)--(0,-1)--(0.2,-0.8)); draw((-0.2,2.8)--(0,3)--(0.2,2.8)); draw((-4.8,-0.2)--(-5,0)--(-4.8,0.2)); draw((2.8,-0.2)--(3,0)--(2.8,0.2)); draw(Q--T); label("$O$",O,SW); label("$P$",P,S); label("$Q$",Q,NE); label("$R$",R,W); label("$T$",T,S); [/asy]
$\text{(A)}\ (-6,0) \qquad \text{(B)}\ (-4,0) \qquad \text{(C)}\ (-2,0) \qquad \text{(D)}\ (2,0) \qquad \text{(E)}\ (4,0)$ | The area of $\square OPQR$ is $2^2 = 4$
The area of $\triangle PQT$ is $\frac{1}{2}bh = \frac{1}{2}\cdot PT\cdot PQ$
If we set the areas equal, the area of $\triangle PQT$ is $4$ . Also, note that $PQ=2$ . Plugging those in, we get:
$4 = \frac{1}{2} \cdot PT \cdot 2$
$PT = 4$
If $PT = 4$ , and $PO = 2$ , then $OT = 2$ , and $T$ must be $2$ units to the left of the origin. This would be $(-2,0)$ , giving answer $\boxed{2,0}$ | C | 2,0 |
38a7e25835bb189882a0f2d9b0d22e35 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_18 | Ana's monthly salary was $$2000$ in May. In June she received a 20% raise. In July she received a 20% pay cut. After the two changes in June and July, Ana's monthly salary was
$\text{(A)}\ 1920\text{ dollars} \qquad \text{(B)}\ 1980\text{ dollars} \qquad \text{(C)}\ 2000\text{ dollars} \qquad \text{(D)}\ 2020\text{ dollars} \qquad \text{(E)}\ 2040\text{ dollars}$ | Notice that a 20% raise translates to Ana's monthly salary multiplied by $(100 + 10)\% = 120\%$ , and a 20% pay cut translates to her salary multiplied by $100\% - 20\% = 80\%$ , so
\[2000 \cdot 120\% = 2400\] \[2400 \cdot 80\% = 1920 \rightarrow \boxed{1920}\] | A | 1920 |
4eea4fac9faa56f94dde5f5b09fe4b40 | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_19 | The pie charts below indicate the percent of students who prefer golf, bowling, or tennis at East Junior High School and West Middle School. The total number of students at East is 2000 and at West, 2500. In the two schools combined, the percent of students who prefer tennis is
[asy] unitsize(18); draw(circle((0,0),4)); draw(circle((9,0),4)); draw((-4,0)--(0,0)--4*dir(352.8)); draw((0,0)--4*dir(100.8)); draw((5,0)--(9,0)--(4*dir(324)+(9,0))); draw((9,0)--(4*dir(50.4)+(9,0))); label("$48\%$",(0,-1),S); label("bowling",(0,-2),S); label("$30\%$",(1.5,1.5),N); label("golf",(1.5,0.5),N); label("$22\%$",(-2,1.5),N); label("tennis",(-2,0.5),N); label("$40\%$",(8.5,-1),S); label("tennis",(8.5,-2),S); label("$24\%$",(10.5,0.5),E); label("golf",(10.5,-0.5),E); label("$36\%$",(7.8,1.7),N); label("bowling",(7.8,0.7),N); label("$\textbf{East JHS}$",(0,-4),S); label("$\textbf{2000 students}$",(0,-5),S); label("$\textbf{West MS}$",(9,-4),S); label("$\textbf{2500 students}$",(9,-5),S); [/asy]
$\text{(A)}\ 30\% \qquad \text{(B)}\ 31\% \qquad \text{(C)}\ 32\% \qquad \text{(D)}\ 33\% \qquad \text{(E)}\ 34\%$ | In the first school, $2000 \cdot 22\% = 2000 \cdot 0.22 = 440$ students prefer tennis.
In the second school, $2500 \cdot 40\% = 2500 \cdot 0.40 = 1000$ students prefer tennis.
In total, $440 + 1000 = 1440$ students prefer tennis out of a total of $2000 + 2500 = 4500$ students
This means $\frac{1440}{4500}\cdot 100\% = \frac{32}{100} \cdot 100\% = 32\%$ of the students in both schools prefer tennis, giving answer $\boxed{32}$ | C | 32 |
93c48a94aece761261b5753f7abe39ee | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_20 | Suppose there is a special key on a calculator that replaces the number $x$ currently displayed with the number given by the formula $1/(1-x)$ . For example, if the calculator is displaying 2 and the special key is pressed, then the calculator will display -1 since $1/(1-2)=-1$ . Now suppose that the calculator is displaying 5. After the special key is pressed 100 times in a row, the calculator will display
$\text{(A)}\ -0.25 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 0.8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 5$ | We look for a pattern, hoping this sequence either settles down to one number, or that it forms a cycle that repeats.
After $1$ press, the calculator displays $\frac{1}{1 - 5} = -\frac{1}{4}$
After $2$ presses, the calculator displays $\frac{1}{1 - (-\frac{1}{4})} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$
After $3$ presses, the calculator displays $\frac{1}{1 - \frac{4}{5}} = \frac{1}{\frac{1}{5}} = 5$
Thus, every three presses, the display will be $5$ . On press $3\cdot 33 = 99$ , the display will be $5$ . One more press will give $-\frac{1}{4}$ , which is answer $\boxed{0.25}$ | A | 0.25 |
dd2b0a1554f4dcbba5ba144e04aba78b | https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_21 | How many subsets containing three different numbers can be selected from the set \[\{ 89,95,99,132, 166,173 \}\] so that the sum of the three numbers is even?
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$ | To have an even sum with three numbers, we must add either $E+O+O$ , or $E + E + E$ , where $O$ represents an odd number, and $E$ represents an even number.
Since there are not three even numbers in the given set, $E+E+E$ is impossible. Thus, we must choose two odd numbers, and one even number.
There are $2$ choices for the even number.
There are $4$ choices for the first odd number. There are $3$ choices for the last odd number. But the order in which we pick these 2 numbers doesn't matter, so this overcounts the pairs of odd numbers by a factor of $2$ . Thus, we have $\frac{4\cdot 3}{2} = 6$ choices for a pair of odd numbers.
In total, there are $2$ choices for an even number, and $6$ choices for the odd numbers, giving a total of $2\cdot 6 = 12$ possible choices for a 3-element set that has an even sum. This is option $\boxed{12}$ | D | 12 |