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1,001
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_1
| 2
|
$(1+11+21+31+41)+(9+19+29+39+49)=$
$\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250$
|
Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us:
\[(1+9)+(11+19)+(21+29)+(31+39)+(41+49),\] which gives us \[10+30+50+70+90 = 250 = \boxed{250}.\]
| 250
|
1,002
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_4
| 1
|
Estimate to determine which of the following numbers is closest to $\frac{401}{.205}$
$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$
|
$401$ is around $400$ and $.205$ is around $.2$ so the fraction is approximately \[\frac{400}{.2}=2000\rightarrow \boxed{2000}\]
| 0
|
1,003
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_5
| 1
|
$-15+9\times (6\div 3) =$
$\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12$
|
We use the order of operations here to get
\begin{align*} -15+9\times (6\div 3) &= -15+9\times 2 \\ &= -15+18 \\ &= 3 \rightarrow \boxed{3}
| 3
|
1,004
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_6
| 1
|
If the markings on the number line are equally spaced, what is the number $\text{y}$
[asy] draw((-4,0)--(26,0),Arrows); for(int a=0; a<6; ++a) { draw((4a,-1)--(4a,1)); } label("0",(0,-1),S); label("20",(20,-1),S); label("y",(12,-1),S); [/asy]
$\text{(A)}\ 3 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 16$
|
Five steps are taken to get from $0$ to $20$ . Each step is of equal size, so each step is $4$ . Three steps are taken from $0$ to $y$ , so $y=3\times 4=12\rightarrow \boxed{12}$
| 12
|
1,005
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_7
| 1
|
If the value of $20$ quarters and $10$ dimes equals the value of $10$ quarters and $n$ dimes, then $n=$
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45$
|
We have \begin{align*} 20(25)+10(10) &= 10(25)+n(10) \\ 600 &= 250+10n \\ 35 &= n \implies \boxed{35}
| 35
|
1,006
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_8
| 1
|
$(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$
|
We use the distributive property to get \[3\times 4+2\times 4+2\times 3 = 26 \rightarrow \boxed{26}\]
| 26
|
1,007
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_8
| 2
|
$(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$
|
Since $\frac12+\frac13+\frac14 > \frac12+\frac14+\frac14 = 1$ , we have \[(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) > 2\times 3\times 4 \times 1 = 24\] The only answer choice greater than $24$ is $\boxed{26}$
| 26
|
1,008
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_10
| 1
|
What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?
$\text{(A)}\ 50^\circ \qquad \text{(B)}\ 120^\circ \qquad \text{(C)}\ 135^\circ \qquad \text{(D)}\ 150^\circ \qquad \text{(E)}\ 165^\circ$
|
The smaller angle makes up $5/12$ of the circle which is the clock. A circle is $360^\circ$ , so the measure of the smaller angle is \[\frac{5}{12}\cdot 360^\circ = 150^\circ \rightarrow \boxed{150}\]
| 150
|
1,009
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_15
| 1
|
The area of the shaded region $\text{BEDC}$ in parallelogram $\text{ABCD}$ is
[asy] unitsize(10); pair A,B,C,D,E; A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0); draw(A--B--C--D--cycle); fill(B--E--D--C--cycle,gray); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S); label("$10$",(9,8),N); label("$6$",(7,0),S); label("$8$",(4,4),W); draw((3,0)--(3,1)--(4,1)); [/asy]
$\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$
|
Let $[ABC]$ denote the area of figure $ABC$
Clearly, $[BEDC]=[ABCD]-[ABE]$ . Using basic area formulas,
Since $AE+ED=BC=10$ and $ED=6$ $AE=4$ and the area of $\triangle ABE$ is $4(4)=16$
Finally, we have $[BEDC]=80-16=64\rightarrow \boxed{64}$
| 64
|
1,010
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_16
| 1
|
In how many ways can $47$ be written as the sum of two primes
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{more than 3}$
|
For $47$ to be written as the sum of two integers , one must be odd and the other must be even. There is only one even prime, namely $2$ , so one of the numbers must be $2$ , making the other $45$
However, $45$ is not prime, so there are no ways to write $47$ as the sum of two primes $\rightarrow \boxed{0}$
| 0
|
1,011
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_17
| 1
|
The number $\text{N}$ is between $9$ and $17$ . The average of $6$ $10$ , and $\text{N}$ could be
$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$
|
We know that $9<N<17$ and we wish to bound $\frac{6+10+N}{3}=\frac{16+N}{3}$
From what we know, we can deduce that $25<N+16<33$ , and thus \[8.\overline{3}<\frac{N+16}{3}<11\]
The only answer choice that falls in this range is choice $\boxed{10}$
| 10
|
1,012
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_20
| 1
|
The figure may be folded along the lines shown to form a number cube. Three number faces come together at each corner of the cube. What is the largest sum of three numbers whose faces come together at a corner?
[asy] draw((0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,1)--(4,1)--(4,0)--(2,0)--(2,-1)--(1,-1)--(1,0)--cycle); draw((1,0)--(1,1)--(2,1)--(2,0)--cycle); draw((3,1)--(3,0)); label("$1$",(1.5,1.25),N); label("$2$",(1.5,.25),N); label("$3$",(1.5,-.75),N); label("$4$",(2.5,.25),N); label("$5$",(3.5,.25),N); label("$6$",(.5,.25),N); [/asy]
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$
|
It is clear that $6$ $5$ , and $4$ will not come together to get a sum of $15$
The faces $6$ $5$ , and $3$ come together at a common vertex, making the maximal sum $6+5+3=14\rightarrow \boxed{14}$
| 14
|
1,013
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_21
| 1
|
Jack had a bag of $128$ apples. He sold $25\%$ of them to Jill. Next he sold $25\%$ of those remaining to June. Of those apples still in his bag, he gave the shiniest one to his teacher. How many apples did Jack have then?
$\text{(A)}\ 7 \qquad \text{(B)}\ 63 \qquad \text{(C)}\ 65 \qquad \text{(D)}\ 71 \qquad \text{(E)}\ 111$
|
First he gives $128\times .25 = 32$ apples to Jill, so he has $128-32=96$ apples left. Then he gives $96\times .25 = 24$ apples to June, so he has $96-24=72$ left.
Finally, he gives one to the teacher, leaving $71\rightarrow \boxed{71}$
| 71
|
1,014
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_22
| 1
|
The letters $\text{A}$ $\text{J}$ $\text{H}$ $\text{S}$ $\text{M}$ $\text{E}$ and the digits $1$ $9$ $8$ $9$ are "cycled" separately as follows and put together in a numbered list:
\[\begin{tabular}[t]{lccc} & & AJHSME & 1989 \\ & & & \\ 1. & & JHSMEA & 9891 \\ 2. & & HSMEAJ & 8919 \\ 3. & & SMEAJH & 9198 \\ & & ........ & \end{tabular}\]
What is the number of the line on which $\text{AJHSME 1989}$ will appear for the first time?
$\text{(A)}\ 6 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 18 \qquad \text{(E)}\ 24$
|
Every $4\text{th}$ line has $1989$ as part of it and every $6\text{th}$ line has $\text{AJHSME}$ as part of it. In order for both to be part of line $n$ $n$ must be a multiple of $4$ and $6$ , the least of which is $\text{lcm}(4,6)=12\rightarrow \boxed{12}$
| 12
|
1,015
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_23
| 1
|
An artist has $14$ cubes, each with an edge of $1$ meter. She stands them on the ground to form a sculpture as shown. She then paints the exposed surface of the sculpture. How many square meters does she paint?
$\text{(A)}\ 21 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 33 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 42$
[asy] draw((0,0)--(2.35,-.15)--(2.44,.81)--(.09,.96)--cycle); draw((.783333333,-.05)--(.873333333,.91)--(1.135,1.135)); draw((1.566666667,-.1)--(1.656666667,.86)--(1.89,1.1)); draw((2.35,-.15)--(4.3,1.5)--(4.39,2.46)--(2.44,.81)); draw((3,.4)--(3.09,1.36)--(2.61,1.4)); draw((3.65,.95)--(3.74,1.91)--(3.29,1.94)); draw((.09,.96)--(.76,1.49)--(.71,1.17)--(2.2,1.1)--(3.6,2.2)--(3.62,2.52)--(4.39,2.46)); draw((.76,1.49)--(.82,1.96)--(2.28,1.89)--(2.2,1.1)); draw((2.28,1.89)--(3.68,2.99)--(3.62,2.52)); draw((1.455,1.135)--(1.55,1.925)--(1.89,2.26)); draw((2.5,2.48)--(2.98,2.44)--(2.9,1.65)); draw((.82,1.96)--(1.55,2.6)--(1.51,2.3)--(2.2,2.26)--(2.9,2.8)--(2.93,3.05)--(3.68,2.99)); draw((1.55,2.6)--(1.59,3.09)--(2.28,3.05)--(2.2,2.26)); draw((2.28,3.05)--(2.98,3.59)--(2.93,3.05)); draw((1.59,3.09)--(2.29,3.63)--(2.98,3.59)); [/asy]
|
We can consider the contributions of the sides of the three layers and the tops of the layers separately.
Layer $n$ (counting from the top starting at $1$ ) has $4$ side faces each with $n$ unit squares, so the sides of the pyramid contribute $4+8+12=24$ for the surface area.
The tops of the layers when combined form the same arrangement of unit cubes as the bottom of the pyramid, which is a $3\times 3$ square, hence this contributes $9$ for the surface area.
You do not have to count the side underneath, since it is not exposed.
Thus, the artist paints $24+9=33 \rightarrow \boxed{33}$ square meters.
| 33
|
1,016
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_2
| 1
|
The product $8\times .25\times 2\times .125 =$
$\text{(A)}\ \frac18 \qquad \text{(B)}\ \frac14 \qquad \text{(C)}\ \frac12 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$
|
Converting the decimals to fractions , this is \begin{align*} 8\times \frac{1}{4} \times 2\times \frac{1}{8} &= \frac{8\times 2}{4\times 8} \\ &= \frac{16}{32} \\ &= \frac{1}{2} \rightarrow \boxed{12}
| 12
|
1,017
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_4
| 1
|
The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$
[asy] unitsize(12); //Force a white background in middle even when transparent fill((3,1)--(12,1)--(12,4)--(3,4)--cycle,white); //Black Squares, Gray Border (blends better than white) for(int a=0; a<7; ++a) { filldraw((2a,0)--(2a+1,0)--(2a+1,1)--(2a,1)--cycle,black,gray); } for(int b=7; b<15; ++b) { filldraw((b,14-b)--(b+1,14-b)--(b+1,15-b)--(b,15-b)--cycle,black,gray); } for(int c=1; c<7; ++c) { filldraw((c,c)--(c+1,c)--(c+1,c+1)--(c,c+1)--cycle,black,gray); } filldraw((6,4)--(7,4)--(7,5)--(6,5)--cycle,black,gray); filldraw((7,5)--(8,5)--(8,6)--(7,6)--cycle,black,gray); filldraw((8,4)--(9,4)--(9,5)--(8,5)--cycle,black,gray); //White Squares, Black Border filldraw((7,4)--(8,4)--(8,5)--(7,5)--cycle,white,black); for(int a=0; a<7; ++a) { filldraw((2a+1,0)--(2a+2,0)--(2a+2,1)--(2a+1,1)--cycle,white,black); } for(int b=9; b<15; ++b) { filldraw((b-1,14-b)--(b,14-b)--(b,15-b)--(b-1,15-b)--cycle,white,black); } for(int c=1; c<7; ++c) { filldraw((c+1,c)--(c+2,c)--(c+2,c+1)--(c+1,c+1)--cycle,white,black); } label("same",(6.3,2.45),N); label("pattern here",(7.5,1.4),N); [/asy]
|
It is simple to notice that in each and every row, there is always one more black square than the white squares. Since there are $8$ rows, there are $8$ more black squares than the white squares. $8\rightarrow \boxed{8}$
| 8
|
1,018
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_5
| 1
|
If $\angle \text{CBD}$ is a right angle , then this protractor indicates that the measure of $\angle \text{ABC}$ is approximately
[asy] unitsize(36); pair A,B,C,D; A=3*dir(160); B=origin; C=3*dir(110); D=3*dir(20); draw((1.5,0)..(0,1.5)..(-1.5,0)); draw((2.5,0)..(0,2.5)..(-2.5,0)--cycle); draw(A--B); draw(C--B); draw(D--B); label("O",(-2.5,0),W); label("A",A,W); label("B",B,S); label("C",C,W); label("D",D,E); label("0",(-1.8,0),W); label("20",(-1.7,.5),NW); label("160",(1.6,.5),NE); label("180",(1.7,0),E); [/asy]
$\text{(A)}\ 20^\circ \qquad \text{(B)}\ 40^\circ \qquad \text{(C)}\ 50^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 120^\circ$
|
We have that $20^{\circ}+\angle ABD +\angle CBD=160^{\circ}$ , or $\angle ABD +\angle CBD=140^{\circ}$ . Since $\angle CBD$ is a right angle, we have $\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \boxed{50}$
| 50
|
1,019
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_12
| 1
|
Suppose the estimated $20$ billion dollar cost to send a person to the planet Mars is shared equally by the $250$ million people in the U.S. Then each person's share is
$\text{(A)}\ 40\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 80\text{ dollars} \qquad \text{(D)}\ 100\text{ dollars} \qquad \text{(E)}\ 125\text{ dollars}$
|
We want the cost per person, which is \begin{align*} \frac{20\text{ billion}}{250\text{ million}} &= \frac{20000\text{ million}}{250\text{ million}} \\ &= 80 \rightarrow \boxed{80}
| 80
|
1,020
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_13
| 1
|
If rose bushes are spaced about $1$ foot apart, approximately how many bushes are needed to surround a circular patio whose radius is $12$ feet?
$\text{(A)}\ 12 \qquad \text{(B)}\ 38 \qquad \text{(C)}\ 48 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 450$
|
The circumference of the patio is $2\cdot 12\cdot \pi =24\pi \approx 75.398$ . Since the bushes are spaced $1$ foot apart, about $75\rightarrow \boxed{75}$ are needed.
| 75
|
1,021
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_14
| 1
|
$\diamondsuit$ and $\Delta$ are whole numbers and $\diamondsuit \times \Delta =36$ . The largest possible value of $\diamondsuit + \Delta$ is
$\text{(A)}\ 12 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20\ \qquad \text{(E)}\ 37$
|
Since it doesn't take too long, we can just make a table with all the possible values of the sum \[\begin{array}{|c|c|c|} \multicolumn{3}{}{} \\ \hline \diamondsuit & \Delta & \diamondsuit + \Delta \\ \hline 36 & 1 & 37 \\ \hline 18 & 2 & 20 \\ \hline 12 & 3 & 15 \\ \hline 9 & 4 & 13 \\ \hline 6 & 6 & 12 \\ \hline \end{array}\]
Clearly $37\rightarrow \boxed{37}$ is the largest.
| 37
|
1,022
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_16
| 1
|
Placing no more than one $\text{X}$ in each small square , what is the greatest number of $\text{X}$ 's that can be put on the grid shown without getting three $\text{X}$ 's in a row vertically, horizontally, or diagonally?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
[asy] for(int a=0; a<4; ++a) { draw((a,0)--(a,3)); } for(int b=0; b<4; ++b) { draw((0,b)--(3,b)); } [/asy]
|
By the Pigeonhole Principle , if there are at least $7$ $\text{X}$ 's, then there will be some row with $3$ $\text{X}$ 's. We can put in $6$ by leaving out the three boxes in one of the main diagonals.
$\rightarrow \boxed{6}$
| 6
|
1,023
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_18
| 1
|
The average weight of $6$ boys is $150$ pounds and the average weight of $4$ girls is $120$ pounds. The average weight of the $10$ children is
$\text{(A)}\ 135\text{ pounds} \qquad \text{(B)}\ 137\text{ pounds} \qquad \text{(C)}\ 138\text{ pounds} \qquad \text{(D)}\ 140\text{ pounds} \qquad \text{(E)}\ 141\text{ pounds}$
|
Let the $6$ boys have total weight $S_B$ and let the $4$ girls have total weight $S_G$ . We are given
\begin{align*} \frac{S_B}{6} &= 150 \\ \frac{S_G}{4} &= 120 \end{align*}
We want the average of the $10$ children, which is \[\frac{S_B+S_G}{10}\] From the first two equations , we can determine that $S_B=900$ and $S_G=480$ , so $S_B+S_G=1380$ . Therefore, the average we desire is \[\frac{1380}{10}=138 \rightarrow \boxed{138}\]
| 138
|
1,024
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_19
| 1
|
What is the $100\text{th}$ number in the arithmetic sequence $1,5,9,13,17,21,25,...$
$\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405$
|
To get from the $1^\text{st}$ term of an arithmetic sequence to the $100^\text{th}$ term, we must add the common difference $99$ times. The first term is $1$ and the common difference is $5-1=9-5=13-9=\cdots = 4$ , so the $100^\text{th}$ term is \[1+4(99)=397 \rightarrow \boxed{397}\]
| 397
|
1,025
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_19
| 2
|
What is the $100\text{th}$ number in the arithmetic sequence $1,5,9,13,17,21,25,...$
$\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405$
|
Alternatively you could create an equation for the arithmetic sequence: $a_{n}=a_{1}+4(n-1)=1+4n-4=4n-3$
$a_{100}=4(100)-3=397$ , or $\boxed{397}$
| 397
|
1,026
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_20
| 1
|
The glass gauge on a cylindrical coffee maker shows that there are $45$ cups left when the coffee maker is $36\%$ full. How many cups of coffee does it hold when it is full?
$\text{(A)}\ 80 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 130 \qquad \text{(E)}\ 262$
[asy] draw((5,0)..(0,-1.3)..(-5,0)); draw((5,0)--(5,10)); draw((-5,0)--(-5,10)); draw(ellipse((0,10),5,1.3)); draw(circle((.3,1.3),.4)); draw((-.1,1.7)--(-.1,7.9)--(.7,7.9)--(.7,1.7)--cycle); fill((-.1,1.7)--(-.1,4)--(.7,4)--(.7,1.7)--cycle,black); draw((-2,11.3)--(2,11.3)..(2.6,11.9)..(2,12.2)--(-2,12.2)..(-2.6,11.9)..cycle); [/asy]
|
Let the amount of coffee the maker will hold when full be $x$ . Then, \[.36x=45 \Rightarrow x=125 \rightarrow \boxed{125}\]
| 125
|
1,027
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_21
| 1
|
A fifth number, $n$ , is added to the set $\{ 3,6,9,10 \}$ to make the mean of the set of five numbers equal to its median . The number of possible values of $n$ is
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ \text{more than }4$
|
The possible medians after $n$ is added are $6$ $n$ , or $9$ . Now we use casework
Case 1: The median is $6$
In this case, $n<6$ and \[\frac{3+n+6+9+10}{5}=6 \Rightarrow n=2\] so this case contributes $1$
Case 2: The median is $n$
We have $6<n<9$ and \[\frac{3+6+n+9+10}{5}=n \Rightarrow n=7\] so this case also contributes $1$
Case 3: The median is $9$
We have $9<n$ and \[\frac{3+6+9+n+10}{5}=9 \Rightarrow 17\] so this case adds $1$
In all there are $3\rightarrow \boxed{3}$ possible values of $n$
| 3
|
1,028
|
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_23
| 1
|
Maria buys computer disks at a price of $4$ for $$5$ and sells them at a price of $3$ for $$5$ . How many computer disks must she sell in order to make a profit of $$100$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 1200$
|
This is the equivalent of saying she buys $12$ for $$15$ and sells $12$ for $$20$ , so for every dozen disks she sells, she profits $$5$
She needs to profit $$100$ , so she needs to sell $\frac{100}{5}=20$ dozen disks, which is $240\rightarrow \boxed{240}$
| 240
|
1,029
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_3
| 1
|
$2(81+83+85+87+89+91+93+95+97+99)=$
$\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$
|
Find that \[(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180\] Which gives us \begin{align*} 2(5 \cdot 180) &= 10 \cdot 180\\ &= 1800 & \text{ Thus \boxed{1800}
| 800
|
1,030
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_3
| 2
|
$2(81+83+85+87+89+91+93+95+97+99)=$
$\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$
|
$2(81+83+85+87+89+91+93+95+97+99)$ Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to $81+99$ $83+97$ $85+95$ $87+93$ $89+91$ $180$ . Since we have $5$ pairs, we multiply $180$ by $5$ to get $900$ . But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get $1800$ , which is $\boxed{1800}$
| 800
|
1,031
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_4
| 1
|
Martians measure angles in clerts. There are $500$ clerts in a full circle. How many clerts are there in a right angle?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 250$
|
The right angle is $1/4$ of the circle, hence it contains $500/4=125\rightarrow \boxed{125}$ clerts.
| 125
|
1,032
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_6
| 1
|
The smallest product one could obtain by multiplying two numbers in the set $\{ -7,-5,-1,1,3 \}$ is
$\text{(A)}\ -35 \qquad \text{(B)}\ -21 \qquad \text{(C)}\ -15 \qquad \text{(D)}\ -1 \qquad \text{(E)}\ 3$
|
To get the smallest possible product, we want to multiply the smallest negative number by the largest positive number. These are $-7$ and $3$ , respectively, and their product is $-21$ , which is $\boxed{21}$
| 21
|
1,033
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_7
| 1
|
The large cube shown is made up of $27$ identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is
$\text{(A)}\ 10 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 22 \qquad \text{(E)}\ 24$
[asy] unitsize(36); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3)); draw((0,3)--(2.2,4.4)--(5.2,4.4)); fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black); fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black); fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); draw((1,3)--(3.2,4.4)); draw((2,3)--(4.2,4.4)); draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666)); draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333)); fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black); fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black); fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black); fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black); fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black); [/asy]
|
Clearly, no unit cube has more than one face painted, so the number of unit cubes with at least one face painted is equal to the number of painted unit squares.
There are $10$ painted unit squares on the half of the cube shown, so there are $10\cdot 2=20$ unit cubes with at least one face painted, thus our answer is $\boxed{20}$
| 20
|
1,034
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_7
| 2
|
The large cube shown is made up of $27$ identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is
$\text{(A)}\ 10 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 22 \qquad \text{(E)}\ 24$
[asy] unitsize(36); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3)); draw((0,3)--(2.2,4.4)--(5.2,4.4)); fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black); fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black); fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); draw((1,3)--(3.2,4.4)); draw((2,3)--(4.2,4.4)); draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666)); draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333)); fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black); fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black); fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black); fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black); fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black); [/asy]
|
Since it says at least one, we can count the number of unpainted cubes, and subtract from 27. There is 1 inner cube, 2 center cubes (see the face with 4 blacks) and 4 edge cubes (see the top two in the center top face), so 7 unpainted. Thus $27 - 7 = 20$ our answer is $\boxed{20}$
| 20
|
1,035
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_9
| 1
|
When finding the sum $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}$ , the least common denominator used is
$\text{(A)}\ 120 \qquad \text{(B)}\ 210 \qquad \text{(C)}\ 420 \qquad \text{(D)}\ 840 \qquad \text{(E)}\ 5040$
|
We want the least common multiple of $2,3,4,5,6,7$ , which is $420$ , or choice $\boxed{420}$
| 420
|
1,036
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_10
| 1
|
$4(299)+3(299)+2(299)+298=$
$\text{(A)}\ 2889 \qquad \text{(B)}\ 2989 \qquad \text{(C)}\ 2991 \qquad \text{(D)}\ 2999 \qquad \text{(E)}\ 3009$
|
We can make use of the distributive property as follows: \begin{align*} 4(299)+3(299)+2(299)+298 &= 4(299)+3(299)+2(299)+1(299)-1 \\ &= (4+3+2+1)(299)-1 \\ &= 10(299)-1 \\ &= 2989 \\ \end{align*}
$\boxed{2989}$
| 989
|
1,037
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_14
| 1
|
A computer can do $10,000$ additions per second. How many additions can it do in one hour?
$\text{(A)}\ 6\text{ million} \qquad \text{(B)}\ 36\text{ million} \qquad \text{(C)}\ 60\text{ million} \qquad \text{(D)}\ 216\text{ million} \qquad \text{(E)}\ 360\text{ million}$
|
There are $3600$ seconds per hour, so we have \begin{align*} \frac{3600\text{ seconds}}{\text{hour}}\cdot \frac{10,000\text{ additions}}{\text{second}} &= \frac{36,000,000\text{ additions}}{\text{hour}} \\ &= 36\text{ million additions per hour} \end{align*}
$\boxed{36}$
| 36
|
1,038
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_15
| 1
|
The sale ad read: "Buy three tires at the regular price and get the fourth tire for 3 dollars." Sam paid 240 dollars for a set of four tires at the sale. What was the regular price of one tire?
$\text{(A)}\ 59.25\text{ dollars} \qquad \text{(B)}\ 60\text{ dollars} \qquad \text{(C)}\ 70\text{ dollars} \qquad \text{(D)}\ 79\text{ dollars} \qquad \text{(E)}\ 80\text{ dollars}$
|
Let the regular price of one tire be $x$ . We have \begin{align*} 3x+3=240 &\Rightarrow 3x=237 \\ &\Rightarrow x=79 \end{align*}
$\boxed{79}$ Good Job!
| 79
|
1,039
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_16
| 1
|
Joyce made $12$ of her first $30$ shots in the first three games of this basketball game, so her seasonal shooting average was $40\%$ . In her next game, she took $10$ shots and raised her seasonal shooting average to $50\%$ . How many of these $10$ shots did she make?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$
|
After the fourth game, she took $40$ shots, $50\%$ of which she made, so she made $40\times .5=20$ shots. Twelve of them were made in the first three games, so in the last game she made $20-12=8$ shots.
$\boxed{8}$
| 8
|
1,040
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_18
| 1
|
Half the people in a room left. One third of those remaining started to dance. There were then $12$ people who were not dancing. The original number of people in the room was
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 72$
|
Let the original number of people in the room be $x$ . Half of them left, so $\frac{x}{2}$ of them are left in the room.
After that, one third of this group is dancing, so $\frac{x}{2}-\frac{1}{3}\left( \frac{x}{2}\right) =\frac{x}{3}$ people are not dancing.
This is given to be $12$ , so \[\frac{x}{3}=12\Rightarrow x=36\]
$\boxed{36}$
| 36
|
1,041
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_18
| 2
|
Half the people in a room left. One third of those remaining started to dance. There were then $12$ people who were not dancing. The original number of people in the room was
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 72$
|
First note that of the $\frac{1}{2}$ people remaining in the room, $\frac{2}{3}$ are not dancing. Therefore $\frac{1}{2}\cdot\frac{2}{3}= \frac{1}{3}$ of the original amount of people in the room is $12$ . The answer is $\boxed{36}$
| 36
|
1,042
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_20
| 1
|
"If a whole number $n$ is not prime, then the whole number $n-2$ is not prime." A value of $n$ which shows this statement to be false is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 23$
|
To show this statement to be false, we need a non-prime value of $n$ such that $n-2$ is prime. Since $13$ and $23$ are prime, they won't prove anything relating to the truth of the statement.
Now we just check the statement for $n=9,12,16$ . If $n=12$ or $n=16$ , then $n-2$ is $10$ or $14$ , which aren't prime. However, $n=9$ makes $n-2=7$ , which is prime, so $n=9$ proves the statement false.
Therefore, the answer is $\boxed{9}$ , 9.
| 9
|
1,043
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_21
| 1
|
Suppose $n^{*}$ means $\frac{1}{n}$ , the reciprocal of $n$ . For example, $5^{*}=\frac{1}{5}$ . How many of the following statements are true?
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$
|
We can just test all of these statements: \begin{align*} 3^*+6^* &= \frac{1}{3}+\frac{1}{6} \\ &= \frac{1}{2} \neq 9^* \\ 6^*-4^* &= \frac{1}{6}-\frac{1}{4} \\ &= \frac{-1}{12} \neq 2^* \\ 2^*\cdot 6^* &= \frac{1}{2}\cdot \frac{1}{6} \\ &= \frac{1}{12} = 12^* \\ 10^* \div 2^* &= \frac{1}{10}\div \frac{1}{2} \\ &= \frac{1}{5} = 5^* \end{align*}
The last two statements are true and the first two aren't, so $\boxed{2}$
| 2
|
1,044
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_23
| 1
|
Assume the adjoining chart shows the $1980$ U.S. population, in millions, for each region by ethnic group. To the nearest percent , what percent of the U.S. Black population lived in the South?
\[\begin{tabular}[t]{c|cccc} & NE & MW & South & West \\ \hline White & 42 & 52 & 57 & 35 \\ Black & 5 & 5 & 15 & 2 \\ Asian & 1 & 1 & 1 & 3 \\ Other & 1 & 1 & 2 & 4 \end{tabular}\]
$\text{(A)}\ 20\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 56\% \qquad \text{(E)}\ 80\%$
|
There are $5+5+15+2=27$ million Blacks living in the U.S. Out of these, $15$ of them live in the South, so the percentage is $\frac{15}{27}\approx \frac{60}{108}\approx 56\%$
$\boxed{56}$
| 56
|
1,045
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_24
| 1
|
A multiple choice examination consists of $20$ questions. The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question. John's score on the examination is $48$ . What is the maximum number of questions he could have answered correctly?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16$
|
Let $c$ be the number of questions correct, $w$ be the number of questions wrong, and $b$ be the number of questions left blank. We are given that \begin{align} c+w+b &= 20 \\ 5c-2w &= 48 \end{align}
Adding equation $(2)$ to double equation $(1)$ , we get \[7c+2b=88\]
Since we want to maximize the value of $c$ , we try to find the largest multiple of $7$ less than $88$ . This is $84=7\times 12$ , so let $c=12$ . Then we have \[7(12)+2b=88\Rightarrow b=2\]
Finally, we have $w=20-12-2=6$ . We want $c$ , so the answer is $12$ , or $\boxed{12}$
| 12
|
1,046
|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_24
| 2
|
A multiple choice examination consists of $20$ questions. The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question. John's score on the examination is $48$ . What is the maximum number of questions he could have answered correctly?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16$
|
If John answered 16 questions correctly, then he answered at most 4 questions incorrectly, giving him at least $16 \cdot 5 - 4 \cdot 2 = 72$ points. Therefore, John did not answer 16 questions correctly. If he answered 12 questions correctly and 6 questions incorrectly (leaving 2 questions unanswered), then he scored $12 \cdot 5 - 6 \cdot 2 = 48$ points. As all other options are less than 12, we conclude that 12 is the most questions John could have answered correctly, and the answer is $\boxed{12}$
| 12
|
1,047
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_3
| 1
|
The smallest sum one could get by adding three different numbers from the set $\{ 7,25,-1,12,-3 \}$ is
$\text{(A)}\ -3 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 21$
|
To find the smallest sum, we just have to find the smallest 3 numbers and add them together.
Obviously, the numbers are $-3, -1, 7$ , and adding them gets us $3$
$\boxed{3}$
| 3
|
1,048
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_4
| 1
|
The product $(1.8)(40.3+.07)$ is closest to
$\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737$
|
Approximating $40.37$ instead of $1.8$ is more effective because larger numbers are less affected by absolute changes (e.g $1001$ is much closer relatively to $1000$ than $2$ is to $1$ ). $74$ is the closest to $72$ , so the answer is $\boxed{74}$
| 74
|
1,049
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_7
| 1
|
How many whole numbers are between $\sqrt{8}$ and $\sqrt{80}$
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$
|
No... of course you're not supposed to know what the square root of 8 is, or the square root of 80. There aren't any formulas, either. Approximation seems like the best strategy.
Clearly it must be true that for any positive integers $a$ $b$ , and $c$ with $a>b>c$ \[\sqrt{a}>\sqrt{b}>\sqrt{c}\]
If we let $a=9$ $b=8$ , and $c=4$ , then we get \[\sqrt{9}>\sqrt{8}>\sqrt{4}\] \[3>\sqrt{8}>2\]
Therefore, the smallest whole number between $\sqrt{8}$ and $\sqrt{80}$ is $3$
Similarly, if we let $a=81$ $b=80$ , and $c=64$ , we get \[\sqrt{81}>\sqrt{80}>\sqrt{64}\] \[9>\sqrt{80}>8\]
So $8$ is the largest whole number between $\sqrt{8}$ and $\sqrt{80}$
So we know that we just have to find the number of integers from 3 to 8 inclusive. If we subtract 2 from every number in this set (which doesn't change the number of integers in the set at all), we find that now all we need to do is find the number of integers there are from 1 to 6, which is obviously 6.
$\boxed{6}$
| 6
|
1,050
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_8
| 1
|
In the product shown, $\text{B}$ is a digit. The value of $\text{B}$ is
\[\begin{array}{rr} &\text{B}2 \\ \times& 7\text{B} \\ \hline &6396 \\ \end{array}\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$
|
Note that in any multiplication problem, the only 2 digits that will influence the last digit of the number will be the last digits of each number being multiplied.
So, $2 \times \text{B}$ has a units digit of $6$ , so $\text{B}$ is either $3$ or $8$ . If $\text{B}=3$ , then the product is $32\times 73$ , which is clearly too small, so $\text{B}=8$
$\boxed{8}$
| 8
|
1,051
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_9
| 1
|
Using only the paths and the directions shown, how many different routes are there from $\text{M}$ to $\text{N}$
[asy] draw((0,0)--(3,0),MidArrow); draw((3,0)--(6,0),MidArrow); draw(6*dir(60)--3*dir(60),MidArrow); draw(3*dir(60)--(0,0),MidArrow); draw(3*dir(60)--(3,0),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(6,0),MidArrow); draw(6*dir(60)--5.1961524227066318805823390245176*dir(30),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--3*dir(60),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(3,0),MidArrow); label("M",6*dir(60),N); label("N",(6,0),SE); label("A",3*dir(60),NW); label("B",5.1961524227066318805823390245176*dir(30),NE); label("C",(3,0),S); label("D",(0,0),SW); [/asy]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
|
There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN.
Since A can only go to C or D, which each only have 1 way to get to N each, there are $1+1=2$ ways to get from A to N.
Since B can only go to A, C or N, and A only has 2 ways to get to N, C only has 1 way and to get from B to N is only 1 way, there are $2+1+1=4$ ways to get from B to N.
M can only go to either B or A, A has 2 ways and B has 4 ways, so M has $4+2=6$ ways to get to N.
6 is $\boxed{6}$
| 6
|
1,052
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_11
| 1
|
If $\text{A}*\text{B}$ means $\frac{\text{A}+\text{B}}{2}$ , then $(3*5)*8$ is
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16\qquad \text{(E)}\ 30$
|
We just plug in and evaluate: \begin{align*} (3*5)*8 &= \left( \frac{3+5}{2}\right) *8 \\ &= 4*8 \\ &= \frac{4+8}{2} \\ &= 6 \\ \end{align*}
$\boxed{6}$
| 6
|
1,053
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_13
| 1
|
The perimeter of the polygon shown is
$\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 48$
$\text{(E)}\ \text{cannot be determined from the information given}$
|
You might have not seen this coming but there is a very simple way to do this. If we try to make a rectangle out of this, we have to take out both of the lines that are taking out part of the rectangle we want to make. but now we see that to finish the rectangle, we have to use those same irregular lines!
[asy] unitsize(12); draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("$6$",(0,3),W); label("$8$",(4,6),N); draw((8,3)--(8,0)--(2.7,0),dashed); [/asy]
So all we have to do is find the perimeter of the shape as if it would be a rectangle. After that, we get $\boxed{28}$
| 28
|
1,054
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_18
| 1
|
A rectangular grazing area is to be fenced off on three sides using part of a $100$ meter rock wall as the fourth side. Fence posts are to be placed every $12$ meters along the fence including the two posts where the fence meets the rock wall. What is the fewest number of posts required to fence an area $36$ m by $60$ m?
[asy] unitsize(12); draw((0,0)--(16,12)); draw((10.66666,8)--(6.66666,13.33333)--(1.33333,9.33333)--(5.33333,4)); label("WALL",(7,4),SE); [/asy]
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$
|
Since we want to minimize the amount of fence that we use, we should have the longer side of the rectangle have one side as the wall. The grazing area is a $36$ m by $60$ m rectangle, so the $60$ m side should be parallel to the wall. That means the two fences perpendicular to the wall are $36$ m. We can start by counting $60\div12+1$ on the $60$ m fence (since we also count the $0$ m post). Next, we have the two $36$ m fences. There are a total of $36\div12+1-1$ fences on that side since the $0$ m and $60$ m fence posts are also part of the $36$ m fences. So we have $6+2\cdot3=12$ minimum fence posts needed to box a $36$ m by $60$ m grazing area, or answer $\boxed{12}$
| 12
|
1,055
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_20
| 1
|
The value of the expression $\frac{(304)^5}{(29.7)(399)^4}$ is closest to
$\text{(A)}\ .003 \qquad \text{(B)}\ .03 \qquad \text{(C)}\ .3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 30$
|
\[\frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}\] Which is closest to $3\rightarrow\boxed{3}$
| 3
|
1,056
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_21
| 1
|
Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined. How many of the resulting figures can be folded into a topless cubical box?
[asy] draw((1,0)--(2,0)--(2,5)--(1,5)--cycle); draw((0,1)--(3,1)--(3,4)--(0,4)--cycle); draw((0,2)--(4,2)--(4,3)--(0,3)--cycle); draw((1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(2,3)--(2,4)--(1,4)--cycle,linewidth(.7 mm)); label("A",(1.5,4.2),N); label("B",(.5,3.2),N); label("C",(2.5,3.2),N); label("D",(.5,2.2),N); label("E",(3.5,2.2),N); label("F",(.5,1.2),N); label("G",(2.5,1.2),N); label("H",(1.5,.2),N); [/asy]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
|
The four squares we already have assemble nicely into four sides of the cube. Let the central one be the bottom, and fold the other three upwards to get the front, right, and back side. Currently, our box is missing its left side and its top side. We have to count the possibilities that would fold to one of these two places.
In total, there are $6\rightarrow\boxed{6}$ good possibilities.
| 6
|
1,057
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_21
| 2
|
Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined. How many of the resulting figures can be folded into a topless cubical box?
[asy] draw((1,0)--(2,0)--(2,5)--(1,5)--cycle); draw((0,1)--(3,1)--(3,4)--(0,4)--cycle); draw((0,2)--(4,2)--(4,3)--(0,3)--cycle); draw((1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(2,3)--(2,4)--(1,4)--cycle,linewidth(.7 mm)); label("A",(1.5,4.2),N); label("B",(.5,3.2),N); label("C",(2.5,3.2),N); label("D",(.5,2.2),N); label("E",(3.5,2.2),N); label("F",(.5,1.2),N); label("G",(2.5,1.2),N); label("H",(1.5,.2),N); [/asy]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
|
Fold the four squares into the four sides of a cube. Then, there are six edges "open" (for lack of better term). For each open edge, we can add a square/side, so the answer is $6\rightarrow\boxed{6}$
| 6
|
1,058
|
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_23
| 1
|
The large circle has diameter $\text{AC}$ . The two small circles have their centers on $\text{AC}$ and just touch at $\text{O}$ , the center of the large circle. If each small circle has radius $1$ , what is the value of the ratio of the area of the shaded region to the area of one of the small circles?
[asy] pair A=(-2,0), O=origin, C=(2,0); path X=Arc(O,2,0,180), Y=Arc((-1,0),1,180,0), Z=Arc((1,0),1,180,0), M=X..Y..Z..cycle; filldraw(M, black, black); draw(reflect(A,C)*M); draw(A--C, dashed); label("A",A,W); label("C",C,E); label("O",O,SE); dot((-1,0)); dot(O); dot((1,0)); label("$1$",(-.5,0),N); label("$1$",(1.5,0),N); [/asy]
$\text{(A)}\ \text{between }\frac{1}{2}\text{ and 1} \qquad \text{(B)}\ 1 \qquad \text{(C)}\ \text{between 1 and }\frac{3}{2}$
$\text{(D)}\ \text{between }\frac{3}{2}\text{ and 2} \qquad \text{(E)}\ \text{cannot be determined from the information given}$
|
The small circle has radius $1$ , thus its area is $\pi$
The large circle has radius $2$ , thus its area is $4\pi$
The area of the semicircle above $AC$ is then $2\pi$
The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is $\pi$ . This means that the area of the shaded part is $2\pi-\pi=\pi$ .
This is equal to the area of a small circle, hence the correct answer is $\boxed{1}$
| 1
|
1,059
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_1
| 1
|
[katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex]
|
By the associative property , we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{1}[/katex]
| 1
|
1,060
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_1
| 2
|
[katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex]
|
Notice that the $9 \times 11$ in the denominator of the first fraction cancels with the same term in the second fraction, the $7$ s in the numerator and denominator of the second fraction cancel, and the $3 \times 5$ in the numerator of the first fraction cancels with the same term in the denominator second fraction. Then everything in the expression cancels, leaving us with $\boxed{1}$
| 1
|
1,061
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2
| 1
|
$90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$
|
To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].
[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.
Rearranging the numbers so each pair sums up to 10, we have:
[mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{945}[/mathjax].
| 945
|
1,062
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2
| 2
|
$90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$
|
We can express each of the terms as a difference from $100$ and then add the negatives using $\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n$ to get the answer. \begin{align*} (100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\ &= 1000 - 55\\ &= \boxed{945}
| 945
|
1,063
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2
| 3
|
$90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$
|
Instead of breaking the sum then rearranging, we can rearrange directly: \begin{align*} 90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ &= 189+189+189+189+189 \\ &= \boxed{945}
| 945
|
1,064
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2
| 4
|
$90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$
|
The finite arithmetic sequence formula states that the sum in the sequence is equal to $\frac{n}{2}\cdot(a_1+a_n)$ where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.
Applying the formula, we have: \[\frac{10}{2}\cdot(90+99)=\boxed{945}\]
| 945
|
1,065
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2
| 5
|
$90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$
|
The expression is equal to the sum of integers from $1$ to $99$ minus the sum of integers from $1$ to $89$ , so it is equal to $\frac{99(100)}{2} - \frac{89(90)}{2} = 4950 - 4005 = \boxed{945}$
| 945
|
1,066
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_3
| 1
|
$\frac{10^7}{5\times 10^4}=$
$\text{(A)}\ .002 \qquad \text{(B)}\ .2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$
|
We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: \[\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}\]
We know that $10^3 = 10 \times 10 \times 10$ , so
\begin{align*} \frac{10^3}{5} &= \frac{10\times 10\times 10}{5} \\ &= 2\times 10\times 10 \\ &= 200 \\ \end{align*}
So the answer is $\boxed{200}$
| 200
|
1,067
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_4
| 1
|
The area of polygon $ABCDEF$ , in square units, is
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74$
[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]
|
[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((2,4)--(6,4),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(6,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]
Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we do know how to find the area of.
If we continue segment $\overline{FE}$ until it reaches the right side at $G$ , we create two rectangles - one on the top and one on the bottom.
We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of $ABGF$ is $6\times5 = 30$ . For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other.
Note that $GC+GB=9$ , and $GB=AF=5$ , so we must have \[GC+5=9\Rightarrow GC=4\]
The area of the bottom rectangle is then \[(DC)(GC)=4\times 4=16\]
Finally, we just add the areas of the rectangles together to get $16 + 30 = 46$
$\boxed{46}$
| 46
|
1,068
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_4
| 2
|
The area of polygon $ABCDEF$ , in square units, is
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74$
[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]
|
[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((0,4)--(0,0),dashed); draw((0,0)--(2,0),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(0,0),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]
Let $\langle ABCDEF \rangle$ be the area of polygon $ABCDEF$ . Also, let $G$ be the intersection of $DC$ and $AF$ when both are extended.
Clearly, \[\langle ABCDEF \rangle = \langle ABCG \rangle - \langle GFED \rangle\]
Since $AB=6$ and $BC=9$ $\langle ABCG \rangle =6\times 9=54$
To compute the area of $GFED$ , note that \[AB=GD+DC\] \[BC=GF+FA\]
We know that $AB=6$ $DC=4$ $BC=9$ , and $FA=5$ , so \[6=GD+4\Rightarrow GD=2\] \[9=GF+5\Rightarrow GF=4\]
Thus $\langle GFED \rangle = 4\times 2=8$
Finally, we have \begin{align*} \langle ABCDEF \rangle &= \langle ABCG \rangle - \langle GFED \rangle \\ &= 54-8 \\ &= 46 \\ \end{align*}
This is answer choice $\boxed{46}$
| 46
|
1,069
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_6
| 1
|
A ream of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high?
$\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250$
|
We could solve the first equation for the thickness of one sheet of paper, and divide into the 2nd equation (which is one way to do the problem), but there are other ways, too.
Let's say that $500\text{ sheets}=5\text{ cm}\Rightarrow \frac{500 \text{ sheets}}{5 \text{ cm}} = 1$ . So by multiplying $7.5 \text{ cm}$ by this fraction, we SHOULD get the number of sheets in 7.5 cm. Solving gets
\begin{align*} \frac{7.5 \times 500}{5} &= 7.5 \times 100 \\ &= 750 \text{ sheets} \\ \end{align*}
$750$ is $\boxed{750}$
| 750
|
1,070
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_6
| 2
|
A ream of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high?
$\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250$
|
We can set up a direct proportion relating the amount of sheets to the thickness because according to the problem, all the papers have the same thickness.
Our proportion is \[\frac{5}{500}=\frac{7.5}{x}\] where $x$ is the number we are looking for.
Next, we cross-multiply to get $5x=500 \times 7.5$ so $x=750$ which is $\boxed{750}$
| 750
|
1,071
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_7
| 1
|
A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is
[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]
$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$
|
The best way to solve this problem is to find patterns and to utilize them to our advantage. For example, we can't really do anything without knowing how many squares there are in the 37th row. But who wants to continue the diagram for 37 rows? And what if the problem said 100,000th row? It'll still be possible - but not if your method is to continue the diagram...
So hopefully there's a pattern. We find a pattern by noticing what is changing from row 1 to row 2. Basically, for the next row, we are just adding 2 squares (1 on each side) to the number of squares we had in the previous row. So each time we're adding 2. So how can we find $N$ , if $N$ is the number of squares in the $a^\text{th}$ row of this diagram? We can't just say that $N = 1 + 2a$ , because it doesn't work for the first row. But since 1 is the first term, we have to EXCLUDE the first term, meaning that we must subtract 1 from a. Thus, $N = 1 + 2\times(a - 1) = 2a - 1$ . So in the 37th row we will have $2 \times 37 - 1 = 74 - 1 = 73$
You may now be thinking - aha, we're finished. But we're only half finished. We still need to find how many black squares there are in these 73 squares. Well let's see - they alternate white-black-white-black... but we can't divide by two - there aren't exactly as many white squares as black squares... there's always 1 more white square... aha! If we subtract 1 from the number of squares (1 white square), we will have exactly 2 times the number of black squares.
Thus, the number of black squares is $\frac{73 - 1}{2} = \frac{72}{2} = 36$
36 is $\boxed{36}$
| 36
|
1,072
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_7
| 2
|
A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is
[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]
$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$
|
Note that each row adds one black and one white square to the end of the previous one, and then shifts the new row over a little. Since the first row had no black squares, the number of black squares in any row is one less than the row number.
Now that this has been established, we just have $37-1=36\rightarrow \boxed{36}$
| 36
|
1,073
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_7
| 3
|
A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is
[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]
$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$
|
We can easily spot the pattern. In the first row, there are $0$ black squares. In the second row, there is $1$ black square. So the pattern is that in order to find the number of black squares in a row, you need to subtract $1$ from the row number. Therefore, the number of black squares in the $37\text{th}$ row we subtract 1. $37-1=36\rightarrow \boxed{36}$
| 36
|
1,074
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_8
| 1
|
If $a = - 2$ , the largest number in the set $\{ - 3a, 4a, \frac {24}{a}, a^2, 1\}$ is
$\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1$
|
Since all the numbers are small, we can just evaluate the set to be \[\{ (-3)(-2), 4(-2), \frac{24}{-2}, (-2)^2, 1 \}= \{ 6, -8, -12, 4, 1 \}\]
The largest number is $6$ , which corresponds to $-3a$
$\boxed{3}$
| 3
|
1,075
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_15
| 1
|
How many whole numbers between $100$ and $400$ contain the digit $2$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148$
|
This is a very common type of counting problem that you'll see quite often. Doing this the simple way would take too long, and might even make lots of mistakes.
If you ever learned about complementary counting , this would be the best time to utilize it. Instead of counting how many DO have 2's, why don't we count how many that DON'T?
So let's find the number of numbers. Obviously, we'd start by subtracting 100 from 400, getting us 300, but we're not done. Since just subtracting includes the number 400, we must subtract one (because 400 isn't allowed - it says between), getting us 299.
So how many numbers are there that DON'T have a 2? Well, we have 2 possibilities for the hundreds digit (1, 3, note that 2 is not allowed), 9 possibilities for the tens digit (1, 3, 4, 5, ... , 9, 0), and 9 possibilities for the ones digit. $2 \times 9 \times 9 = 162$ . However, one of the numbers we counted is $100$ , which isn't allowed, so there are $162-1=161$ numbers without a 2.
Since there are 299 numbers in total and 161 that DON'T have any 2's, $299 - 161 = 138$ numbers WILL have at least one two.
$\boxed{138}$
| 138
|
1,076
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_15
| 2
|
How many whole numbers between $100$ and $400$ contain the digit $2$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148$
|
As in the previous solution, get rid of the $400$ to make things easier. This gives us the numbers from $100$ to $399$
Let $A$ be the event that the first digit is $2$ $B$ be the event that the second digit is $2$ , and $C$ be the event that the third digit is $2$ . Then, PIE says that our answer will be $|A|+|B|+|C|-|A\cup B|-|A\cup C|-|B\cup C|+|A\cup B\cup C|$
We have that $|A|$ is just $1\times10\times10=100$ $|B|=3\times1\times10=30$ , and $|C|=3\times10\times1=30$
Next, $|A\cup B|$ is just having something in the form $22\_$ , so there are $10$ ways. Similarly, $|A\cup C|$ means that we have $2\_2$ , so there are again $10$ ways. Finally, $|B\cup C|$ means that our number is like $\_22$ , so there are $3$ ways.
Finally, $|A\cup B\cup C|$ counts the number of three digit numbers with all three digits $2$ , which there is only $1$ of: $222$
Putting this together, our answer will be $100+30+30-10-10-3+1=\boxed{138}$
| 138
|
1,077
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_16
| 1
|
The ratio of boys to girls in Mr. Brown's math class is $2:3$ . If there are $30$ students in the class, how many more girls than boys are in the class?
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 10$
|
Let the number of boys be $2x$ . It follows that the number of girls is $3x$ . These two values add up to $30$ students, so \[2x+3x=5x=30\Rightarrow x=6\]
The difference between the number of girls and the number of boys is $3x-2x=x$ , which is $6$ , so the answer is $\boxed{6}$
| 6
|
1,078
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_17
| 1
|
If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$ , then your score on the seventh test was
$\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92$
|
If the average score of the first six is $84$ , then the sum of those six scores is $6\times 84=504$
The average score of the first seven is $85$ , so the sum of the seven is $7\times 85=595$
Taking the difference leaves us with just the seventh score, which is $595-504=91$ , so the answer is $\boxed{91}$
| 91
|
1,079
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_17
| 2
|
If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$ , then your score on the seventh test was
$\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92$
|
Let's remove the condition that the average of the first seven tests is $85$ , and say the 7th test score was a $0$ . Then, the average of the first seven tests would be \[\frac{6\times 84}{7}=72\]
If we increase the seventh test score by $7$ , the average will increase by $\frac{7}{7}=1$ . We need the average to increase by $85-72=13$ , so the seventh test score is $7\times 13=91$ more than $0$ , which is clearly $91$ . This is choice $\boxed{91}$
| 91
|
1,080
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_19
| 1
|
If the length and width of a rectangle are each increased by $10\%$ , then the perimeter of the rectangle is increased by
$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$
|
Let the width be $w$ and the length be $l$ . Then, the original perimeter is $2(w+l)$
After the increase, the new width and new length are $1.1w$ and $1.1l$ , so the new perimeter is $2(1.1w+1.1l)=2.2(w+l)$
Therefore, the percent change is \begin{align*} \frac{2.2(w+l)-2(w+l)}{2(w+l)} &= \frac{.2(w+l)}{2(w+l)} \\ &= \frac{.2}{2} \\ &= 10\% \\ \end{align*}
$\boxed{10}$
| 10
|
1,081
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_19
| 2
|
If the length and width of a rectangle are each increased by $10\%$ , then the perimeter of the rectangle is increased by
$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$
|
Assume WLOG that the rectangle is a square with length $10$ and width $10$ . Thus, the square has a perimeter of $40$ . Increasing the length and width by $10\%$ will increase the dimensions of the square to 11x11. Thus, the new square's perimeter is $44$ , and because $44$ is $110\%$ of $40$ , our answer is $\boxed{10}$
| 10
|
1,082
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_21
| 1
|
Mr. Green receives a $10\%$ raise every year. His salary after four such raises has gone up by what percent?
$\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\%$
|
Assume his salary is originally $100$ dollars. Then, in the next year, he would have $110$ dollars, and in the next, he would have $121$ dollars. The next year he would have $133.1$ dollars and in the final year, he would have $146.41$ . As the total increase is greater than $45\%$ , the answer is $\boxed{45}$
| 45
|
1,083
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_23
| 1
|
King Middle School has $1200$ students. Each student takes $5$ classes a day. Each teacher teaches $4$ classes. Each class has $30$ students and $1$ teacher. How many teachers are there at King Middle School?
$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50$
|
If each student has $5$ classes, and there are $1200$ students, then they have a total of $5\times 1200=6000$ classes among them.
Each class has $30$ students, so there must be $\frac{6000}{30}=200$ classes. Each class has $1$ teacher, so the teachers have a total of $200$ classes among them.
Each teacher teaches $4$ classes, so if there are $t$ teachers, they have $4t$ classes among them. This was found to be $200$ , so \[4t=200\Rightarrow t=50\]
This is answer choice $\boxed{50}$
| 50
|
1,084
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_24
| 1
|
In a magic triangle, each of the six whole numbers $10-15$ is placed in one of the circles so that the sum, $S$ , of the three numbers on each side of the triangle is the same. The largest possible value for $S$ is
[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir(60)); draw(circle(14*dir(60),1)); draw((1,0)--(6,0)); draw(circle((7,0),1)); draw((8,0)--(13,0)); draw(circle((14,0),1)); draw(circle((10.5,6.0621778264910705273460621952706),1)); draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176)); draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788)); [/asy]
$\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40$
|
Let the number in the top circle be $a$ and then $b$ $c$ $d$ $e$ , and $f$ , going in clockwise order. Then, we have \[S=a+b+c\] \[S=c+d+e\] \[S=e+f+a\]
Adding these equations together, we get
\begin{align*} 3S &= (a+b+c+d+e+f)+(a+c+e) \\ &= 75+(a+c+e) \\ \end{align*}
where the last step comes from the fact that since $a$ $b$ $c$ $d$ $e$ , and $f$ are the numbers $10-15$ in some order, their sum is $10+11+12+13+14+15=75$
The left hand side is divisible by $3$ and $75$ is divisible by $3$ , so $a+c+e$ must be divisible by $3$ . The largest possible value of $a+c+e$ is then $15+14+13=42$ , and the corresponding value of $S$ is $\frac{75+42}{3}=39$ , which is choice $\boxed{39}$
| 39
|
1,085
|
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_25
| 1
|
Five cards are lying on a table as shown.
\[\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\ \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}\]
|
Using the answer choices, we see that P and Q are logically equivalent (both are non-vowel letters) and so are $4$ and $6$ (both are even numbers). Thus, if one of these is correct, then the other option in its pair must also be correct. There can't be 2 answers. So, the only remaining answer choice is $\boxed{3}$
| 3
|
1,086
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_4
| 1
|
Find the sum of all prime numbers between $1$ and $100$ that are simultaneously $1$ greater than a multiple of $4$ and $1$ less than a multiple of $5$
$\mathrm{(A) \ } 118 \qquad \mathrm{(B) \ }137 \qquad \mathrm{(C) \ } 158 \qquad \mathrm{(D) \ } 187 \qquad \mathrm{(E) \ } 245$
|
Numbers that are $1$ less than a multiple of $5$ all end in $4$ or $9$
No prime number ends in $4$ , since all numbers that end in $4$ are divisible by $2$ . Thus, we are only looking for numbers that end in $9$
Writing down the ten numbers that so far qualify, we get $9, 19, 29, 39, 49, 59, 69, 79, 89, 99$
Crossing off multiples of $3$ gives $19, 29, 49, 59, 79, 89$
Crossing off numbers that are not $1$ more than a multiple of $4$ (in other words, numbers that are $1$ less than a multiple of $4$ , since all numbers are odd), we get:
$29, 49, 89.$
Noting that $49$ is not prime, we have only $29$ and $89$ , which give a sum of $118$ , so the answer is $\boxed{118}$
| 118
|
1,087
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_5
| 1
|
The marked price of a book was 30% less than the suggested retail price. Alice purchased the book for half the marked price at a Fiftieth Anniversary sale. What percent of the suggested retail price did Alice pay?
$\mathrm{(A)\ }25\%\qquad\mathrm{(B)\ }30\%\qquad\mathrm{(C)\ }35\%\qquad\mathrm{(D)\ }60\%\qquad\mathrm{(E)\ }65\%$
|
Without loss of generality, let's assume that the retail price was $100$ USD.
The marked price of the book is $30 \%$ off of $100$ which is equal to $100-100(0.3)=70.$
Half of that marked price is $0.5(70)=35.$
Therefore the percent Alice payed of the suggested retail price is $35/100=\boxed{35}.$
| 35
|
1,088
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_6
| 1
|
What is the sum of the digits of the decimal form of the product $2^{1999}\cdot 5^{2001}$
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10$
|
$2^{1999}\cdot5^{2001}=2^{1999}\cdot5^{1999}\cdot5^{2}=25\cdot10^{1999}$ , a number with the digits "25" followed by 1999 zeros. The sum of the digits in the decimal form would be $2+5=7$ , thus making the answer $\boxed{7}$
| 7
|
1,089
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_7
| 1
|
What is the largest number of acute angles that a convex hexagon can have?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$
|
The sum of the interior angles of a hexagon is $720$ degrees. In a convex polygon, each angle must be strictly less than $180$ degrees.
Six acute angles can only sum to less than $90\cdot 6 = 540$ degrees, so six acute angles could not form a hexagon.
Five acute angles and one obtuse angle can only sum to less than $90\cdot 5 + 180 = 630$ degrees, so these angles could not form a hexagon.
Four acute angles and two obtuse angles can only sum to less than $90\cdot 4 + 180\cdot 2 = 720$ degrees. This is a strict inequality, so these angles could not form a hexagon. (The limiting figure would be four right angles and two straight angles, which would really be a square with two "extra" points on two sides to form the straight angles.)
Three acute angles and three obtuse angles work. For example, if you pick three acute angles of $80$ degrees, the three obtuse angles would be $160$ degrees and give a sum of $80\cdot 3 + 160\cdot 3 = 720$ degrees, which is a genuine hexagon. Thus, the answer is $\boxed{3}$
| 3
|
1,090
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_8
| 1
|
At the end of $1994$ , Walter was half as old as his grandmother. The sum of the years in which they were born was $3838$ . How old will Walter be at the end of $1999$
$\textbf{(A)}\ 48 \qquad \textbf{(B)}\ 49\qquad \textbf{(C)}\ 53\qquad \textbf{(D)}\ 55\qquad \textbf{(E)}\ 101$
|
In $1994$ , if Water is $x$ years old, then Walter's grandmother is $2x$ years old.
This means that Walter was born in $1994 - x$ , and Walter's grandmother was born in $1994 - 2x$
The sum of those years is $3838$ , so we have:
$1994 - x + 1994 - 2x = 3838$
$3988 - 3x = 3838$
$x = 50$
If Walter is $50$ years old in $1994$ , then he will be $55$ years old in $1999$ , thus giving answer $\boxed{55}$
| 55
|
1,091
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_11
| 1
|
The student locker numbers at Olympic High are numbered consecutively beginning with locker number $1$ . The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number $9$ and four centers to label locker number $10$ . If it costs $137.94 to label all the lockers, how many lockers are there at the school?
$\textbf{(A)}\ 2001 \qquad \textbf{(B)}\ 2010 \qquad \textbf{(C)}\ 2100 \qquad \textbf{(D)}\ 2726 \qquad \textbf{(E)}\ 6897$
|
Since all answers are over $2000$ , work backwards and find the cost of the first $1999$ lockers. The first $9$ lockers cost $0.18$ dollars, while the next $90$ lockers cost $0.04\cdot 90 = 3.60$ . Lockers $100$ through $999$ cost $0.06\cdot 900 = 54.00$ , and lockers $1000$ through $1999$ inclusive cost $0.08\cdot 1000 = 80.00$
This gives a total cost of $0.18 + 3.60 + 54.00 + 80.00 = 137.78$ . There are $137.94 - 137.78 = 0.16$ dollars left over, which is enough for $8$ digits, or $2$ more four digit lockers. These lockers are $2000$ and $2001$ , leading to answer $\boxed{2001}$
| 1
|
1,092
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_12
| 1
|
What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions $y=p(x)$ and $y=q(x)$ , each with leading coefficient 1?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
|
The intersections of the two polynomials, $p(x)$ and $q(x)$ , are precisely the roots of the equation $p(x)=q(x) \rightarrow p(x) - q(x) = 0$ . Since the leading coefficients of both polynomials are $1$ , the degree of $p(x) - q(x) = 0$ is at most three, and the maximum point of intersection is three, because a third degree polynomial can have at most three roots. Thus, the answer is $\boxed{3}$
| 3
|
1,093
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_14
| 1
|
Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang $7$ songs, which was more than any other girl, and Mary sang $4$ songs, which was fewer than any other girl. How many songs did these trios sing?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 11$
|
Alina and Tina must sing more than $4$ , but less than $7$ , songs. Therefore, Alina sang $5$ or $6$ songs, and Tina sang $5$ or $6$ songs, with $4$ possible combinations.
However, since every song is a trio, if you add up all the numbers of songs a person sang for all four singers, it must be divisible by $3$ . Thus, $7 + 4 + a + t$ must be divisible by $3$ , and $a+t - 1$ must be divisible by $3$
Since $10 \le a+t \le 12$ by the bounds of $5 \le a, t \le 6$ , we have $9 \le a + t - 1 \le 11$ . Because the middle number must be a multiple of $3$ , we set $a + t - 1 = 9$ , leading to $a + t = 10$ and $a = t = 5$
So, the four girls sang $4, 5, 5$ and $7$ songs in trios. That means there were $\frac{4 + 5 + 5 + 7}{3} = 7$ songs sung, and the answer is $\boxed{7}$
| 7
|
1,094
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_17
| 1
|
Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-19$ , the remainder is $99$ , and when $P(x)$ is divided by $x - 99$ , the remainder is $19$ . What is the remainder when $P(x)$ is divided by $(x-19)(x-99)$
$\mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0$
|
According to the problem statement, there are polynomials $Q(x)$ and $R(x)$ such that $P(x) = Q(x)(x-19) + 99 = R(x)(x-99) + 19$
From the last equality we get $Q(x)(x-19) + 80 = R(x)(x-99)$
The value $x=99$ is a root of the polynomial on the right hand side, therefore it must be a root of the one on the left hand side as well. Substituting, we get $Q(99)(99-19) + 80 = 0$ , from which $Q(99)=-1$ . This means that $99$ is a root of the polynomial $Q(x)+1$ . In other words, there is a polynomial $S(x)$ such that $Q(x)+1 = S(x)(x-99)$
Substituting this into the original formula for $P(x)$ we get \[P(x) = Q(x)(x-19) + 99 = (S(x)(x-99) - 1)(x-19) + 99 =\] \[= S(x)(x-99)(x-19) - (x-19) + 99\]
Therefore when $P(x)$ is divided by $(x-19)(x-99)$ , the remainder is $\boxed{118}$
| 118
|
1,095
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_17
| 2
|
Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-19$ , the remainder is $99$ , and when $P(x)$ is divided by $x - 99$ , the remainder is $19$ . What is the remainder when $P(x)$ is divided by $(x-19)(x-99)$
$\mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0$
|
Since the divisor $(x-19)(x-99)$ is a quadratic, the degree of the remainder is at most linear. We can write $P(x)$ in the form \[P(x) = Q(x)(x-19)(x-99) + cx+d\] where $cx+d$ is the remainder.
By the Remainder Theorem, plugging in $19$ and $99$ gives us a system of equations. \[99c+d = 19\] \[19c+d = 99\]
Solving gives us $c=-1$ and $d = 118$ , thus, our answer is $\boxed{118}$
| 118
|
1,096
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_20
| 1
|
The sequence $a_{1},a_{2},a_{3},\ldots$ satisfies $a_{1} = 19,a_{9} = 99$ , and, for all $n\geq 3$ $a_{n}$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2$
$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$
|
Let $m$ be the arithmetic mean of $a_1$ and $a_2$ . We can then write $a_1=m-x$ and $a_2=m+x$ for some $x$
By definition, $a_3=m$
Next, $a_4$ is the mean of $m-x$ $m+x$ and $m$ , which is again $m$
Realizing this, one can easily prove by induction that $\forall n\geq 3;~ a_n=m$
It follows that $m=a_9=99$ . From $19=a_1=m-x$ we get that $x=80$ . And thus $a_2 = m+x = \boxed{179}$
| 179
|
1,097
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_20
| 2
|
The sequence $a_{1},a_{2},a_{3},\ldots$ satisfies $a_{1} = 19,a_{9} = 99$ , and, for all $n\geq 3$ $a_{n}$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2$
$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$
|
Let $a_1=a$ and $a_2=b$ . Then, $a_3=\frac{a+b}{2}$ $a_4=\frac{a+b+\frac{a+b}{2}}{3}=\frac{a+b}{2},$ and so on.
Since $a_3=a_4$ $a_n=a_3$ for all $n\geq3.$
Hence, $a_9=\frac{a_1+a_2}{2}=\frac{a+b}{2}=99, a+b=198.$ We also know that $a_1=a=19.$
Subtracting $a_1$ from $198,$ we get $b=a_2=\boxed{179}.$
| 179
|
1,098
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_22
| 1
|
The graphs of $y = -|x-a| + b$ and $y = |x-c| + d$ intersect at points $(2,5)$ and $(8,3)$ . Find $a+c$
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18$
|
Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a $45^\circ$ angle, in the second graph they point upwards. One can easily find out that the only way how to get these graphs to intersect in two points is the one depicted below:
Obviously, the maximum of the first graph is achieved when $x=a$ , and its value is $-0+b=b$ . Similarly, the minimum of the other graph is $(c,d)$ . Therefore the two remaining vertices of the area between the graphs are $(a,b)$ and $(c,d)$
As the area has four right angles, it is a rectangle. Without actually computing $a$ and $c$ we can therefore conclude that $a+c=2+8=\boxed{10}$
| 10
|
1,099
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_25
| 1
|
There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ . Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$
|
Multiply out the $7!$ to get
\[5 \cdot 6! = (3 \cdot 4 \cdots 7)a_2 + (4 \cdots 7)a_3 + (5 \cdot 6 \cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .\]
By Wilson's Theorem (or by straightforward division), $a_7 + 7(a_6 + 6a_5 + \cdots) \equiv 5 \cdot 6! \equiv -5 \equiv 2 \pmod{7}$ , so $a_7 = 2$ . Then we move $a_7$ to the left and divide through by $7$ to obtain
\[\frac{5 \cdot 6!-2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.\]
We then repeat this procedure $\pmod{6}$ , from which it follows that $a_6 \equiv 514 \equiv 4 \pmod{6}$ , and so forth. Continuing, we find the unique solution to be $(a_2, a_3, a_4, a_5, a_6, a_7) = (1,1,1,0,4,2)$ (uniqueness is assured by the Division Theorem ). The answer is $9 \Longrightarrow \boxed{9}$
| 9
|
1,100
|
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_25
| 2
|
There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ . Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$
|
We start by multiplying both sides by $7!$ , and we get: \[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\] After doing some guess and check, we find that the answer is $\boxed{9}$
| 9
|
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