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fe9158b8e93c11693e259fb0c7818076 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$ | Let the three numbers be, in increasing order, $9,y,z$
Hence, we have that $9-y=y-z\implies 9+z=2y$
Also, from the second part of information given, we get that
$\frac{9}{y+2}=\frac{y+2}{z+20}\implies 9(z+20)=(y+2)^2\implies y=3(\sqrt{z+20})-2$
Plugging back in...
$9+z=6(\sqrt{z+20})-4\implies (9+z)^2=36(z+20)$
Simplifying, we get that $z^2-10z-551=0$
Applying the quadratic formula, we get that $z=\frac{10\pm \sqrt{2304}}{2}\implies \frac{10\pm48}{2}$
Obviously, in order to minimize the value of $z$ , we have to subtract. Hence, $z=-19$
However, the problem asks for the minimum value of the third term in a geometric progression.
Hence, the answer is $-19+20=\boxed{1}$ | A | 1 |
fe9158b8e93c11693e259fb0c7818076 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$ | Let the arithmetic sequence be $9,9+x,9+2x$ and let the geometric sequence be $9,11+x,29+2x$ . Now, we just try all the solutions. If the last term is $1$ , then $x=-14$ . This gives the geometric sequence $9,-3,1$ which indeed works. The answer is $\boxed{1}$ | A | 1 |
fe9158b8e93c11693e259fb0c7818076 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$ | The terms of the arithmetic progression are 9, $9+d$ , and $9+2d$ for some real number $d$ . The terms of the geometric progression are 9, $11+d$ , and $29+2d$ . Therefore $(11+d)^{2} = 9(29+2d) \quad\text{so}\quad d^{2}+4d-140 = 0.$ Thus $d=10$ or $d=-14$ . The corresponding geometric progressions are $9, 21, 49$ and $9, -3, 1,$ so the smallest possible value for the third term of the geometric progression is $\boxed{1}$ | A | 1 |
fe9158b8e93c11693e259fb0c7818076 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$ | List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is $\boxed{1}$ | A | 1 |
58fabac7ed25a71636dd5ef85daba37a | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_15 | Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
$\mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500$ | Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $\frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a proportion $\frac{100}{x- 150} = \frac{\frac{x}{2} - 100}{150}$ . Cross-multiplying, we get that $x = 350\Longrightarrow\boxed{350}$ | C | 350 |
58fabac7ed25a71636dd5ef85daba37a | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_15 | Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
$\mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500$ | The total distance the girls run between the start and the first meeting is one half of the track length. The total distance they run between the two meetings is the track length. As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting. Thus between the meetings Brenda will run $2\times100=200$ meters. Therefore the length of the track is $150 + 200 = 350$ meters $\Rightarrow\boxed{350}$ | C | 350 |
909fcd06640d729a1ebbf5b7c34b11a2 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_25 | For each integer $n\geq 4$ , let $a_n$ denote the base- $n$ number $0.\overline{133}_n$ . The product $a_4a_5\cdots a_{99}$ can be expressed as $\frac {m}{n!}$ , where $m$ and $n$ are positive integers and $n$ is as small as possible. What is $m$
$\text {(A)} 98 \qquad \text {(B)} 101 \qquad \text {(C)} 132\qquad \text {(D)} 798\qquad \text {(E)}962$ | Note that \[0.\overline{133}_n = \frac{n^2+3n+3}{n^3-1},\] by geometric series.
Thus, we're aiming to find the value of \[\prod_{k=4}^{99} \frac{k^2+3k+3}{k^3 - 1}.\] Expanding the product out, this is equivalent to \[\frac{4^2+3(4)+3}{4^3 - 1} \cdot \frac{5^2+3(5)+3}{5^3 - 1} \cdot \frac{6^2+3(6)+3}{6^3 - 1} \cdot ... \cdot \frac{99^2+3(99)+3}{99^3 - 1}.\] Note that the numerator of the $a$ th fraction and the denominator of the $a+1$ th fraction for $1 \leq a \leq 95$ cancel out to be $\frac{1}{a+3},$ since \[\frac{k^2 + 3k + 3}{(k+1)^3 - 1} = \frac{k^2 + 3k + 3}{k^3 + 3k^2 + 3k} = \frac{1}{k},\] by the binomial theorem on the the denominator of the aforementioned. Since this forms a telescoping series, our product is now equivalent to \[\frac{99^2 + 3(99) + 3}{4^3 - 1} \cdot \prod_{k=4}^{98} \frac{1}{k},\] which, after simplification gives $\frac{6}{98!} \cdot \frac{10101}{63} = \frac{962}{98!},$ giving an answer of $\boxed{962}.$ | null | 962 |
a25fa3f5831facfdacb1c125596ae6e9 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_1 | At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 48 free throws. How many free throws did she make at the first practice?
$(\mathrm {A}) 3\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15$ | Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made $\frac{1}{2} \cdot 48 = 24$ free throws, on the third $12$ , on the second $6$ , and on the first $3 \Rightarrow \mathrm{(A)}$
Because there are five days, or four transformations between days (day 1 $\rightarrow$ day 2 $\rightarrow$ day 3 $\rightarrow$ day 4 $\rightarrow$ day 5), she makes $48 \cdot \frac{1}{2^4} = \boxed{3}$ | A | 3 |
25d25127eebc7736c1c1a343be259bed | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_2 | In the expression $c\cdot a^b-d$ , the values of $a$ $b$ $c$ , and $d$ are $0$ $1$ $2$ , and $3$ , although not necessarily in that order. What is the maximum possible value of the result?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }8\qquad\mathrm{(D)\ }9\qquad\mathrm{(E)\ }10$ | If $a=0$ or $c=0$ , the expression evaluates to $-d<0$ If $b=0$ , the expression evaluates to $c-d\leq 2$ Case $d=0$ remains.
In that case, we want to maximize $c\cdot a^b$ where $\{a,b,c\}=\{1,2,3\}$ . Trying out the six possibilities we get that the greatest is $(a,b,c)=(3,2,1)$ , where $c\cdot a^b=1\cdot 3^2=\boxed{9}$ | D | 9 |
197af6032a14880c75bea118b6b87dc8 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_3 | If $x$ and $y$ are positive integers for which $2^x3^y=1296$ , what is the value of $x+y$
$(\mathrm {A})\ 8 \qquad (\mathrm {B})\ 9 \qquad (\mathrm {C})\ 10 \qquad (\mathrm {D})\ 11 \qquad (\mathrm {E})\ 12$ | $1296 = 2^4 3^4$ and $4+4=\boxed{8}$ | A | 8 |
20fe9535cefbd4bb2dcf3eee5ed5450f | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_5 | On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }7\qquad\mathrm{(D)\ }8\qquad\mathrm{(E)\ }9$ | Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=\frac{7}{10}\cdot(60+d)$ , which solves to $d=140$ , and the sum of digits of $d$ is $\boxed{5}$ | A | 5 |
20fe9535cefbd4bb2dcf3eee5ed5450f | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_5 | On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }7\qquad\mathrm{(D)\ }8\qquad\mathrm{(E)\ }9$ | Each time Isabella exchanges $7$ U.S. dollars, she gets $7$ Canadian dollars and $3$ Canadian dollars extra. Isabella received a total of $60$ Canadian dollars extra, therefore she exchanged $7$ U.S. dollars $\frac{60}{3}=20$ times. Thus $d=7\cdot20=140$ , and the sum of the digits is $\boxed{5}$ | A | 5 |
856ba213af83729e2e2c37f7c1c53516 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_6 | Minneapolis-St. Paul International Airport is $8$ miles southwest of downtown St. Paul and $10$ miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?
$\mathrm{(A)\ }13\qquad\mathrm{(B)\ }14\qquad\mathrm{(C)\ }15\qquad\mathrm{(D)\ }16\qquad\mathrm{(E)\ }17$ | The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs $8$ miles and $10$ miles long. The hypotenuse length is $\sqrt{8^2 + 10^2}\approx12.8$ , and thus the answer is $\boxed{13}$ | A | 13 |
72070bd2fde24d404cb6e4c6d7b04366 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_8 | A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $100$ cans, how many rows does it contain?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }8\qquad\mathrm{(C)\ }9\qquad\mathrm{(D)\ }10\qquad\mathrm{(E)\ }11$ | The sum of the first $n$ odd numbers is $n^2$ . As in our case $n^2=100$ , we have $n=\boxed{10}$ | D | 10 |
9c8a4672262caa2438fad883a795ec00 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_9 | The point $(-3,2)$ is rotated $90^\circ$ clockwise around the origin to point $B$ . Point $B$ is then reflected over the line $x=y$ to point $C$ . What are the coordinates of $C$
$\mathrm{(A)}\ (-3,-2) \qquad \mathrm{(B)}\ (-2,-3) \qquad \mathrm{(C)}\ (2,-3) \qquad \mathrm{(D)}\ (2,3) \qquad \mathrm{(E)}\ (3,2)$ | The entire situation is in the picture below. The correct answer is $\boxed{3,2}$ | E | 3,2 |
4dc77293dd11ac11b7c85b698bc26cec | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_10 | An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$ , with $b>c$ . Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$ , and let $OY$ be the radius of the larger circle that contains $Z$ . Let $a=XZ$ $d=YZ$ , and $e=XY$ . What is the area of the annulus?
[asy] unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label("$O$",O,SW); label("$X$",X,E); label("$Y$",Y,N); label("$Z$",Z,SW); label("$a$",X--Z,N); label("$b$",0.25*X,SE); label("$c$",O--Z,E); label("$d$",Y--Z,W); label("$e$",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]
$\mathrm{(A) \ } \pi a^2 \qquad \mathrm{(B) \ } \pi b^2 \qquad \mathrm{(C) \ } \pi c^2 \qquad \mathrm{(D) \ } \pi d^2 \qquad \mathrm{(E) \ } \pi e^2$ | The area of the large circle is $\pi b^2$ , the area of the small one is $\pi c^2$ , hence the shaded area is $\pi(b^2-c^2)$
From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$ , hence $b^2-c^2=a^2$ and thus the shaded area is $\boxed{2}$ | A | 2 |
4dc77293dd11ac11b7c85b698bc26cec | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_10 | An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$ , with $b>c$ . Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$ , and let $OY$ be the radius of the larger circle that contains $Z$ . Let $a=XZ$ $d=YZ$ , and $e=XY$ . What is the area of the annulus?
[asy] unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label("$O$",O,SW); label("$X$",X,E); label("$Y$",Y,N); label("$Z$",Z,SW); label("$a$",X--Z,N); label("$b$",0.25*X,SE); label("$c$",O--Z,E); label("$d$",Y--Z,W); label("$e$",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]
$\mathrm{(A) \ } \pi a^2 \qquad \mathrm{(B) \ } \pi b^2 \qquad \mathrm{(C) \ } \pi c^2 \qquad \mathrm{(D) \ } \pi d^2 \qquad \mathrm{(E) \ } \pi e^2$ | Set $c=0,$ then the shaded area is just the area of a circle with radius $a,$ which is $\boxed{2}$ | A | 2 |
26e9c0e5f49f4132c671d9a2e2698f97 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_11 | All the students in an algebra class took a $100$ -point test. Five students scored $100$ , each student scored at least $60$ , and the mean score was $76$ . What is the smallest possible number of students in the class?
$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ 14$ | Let the number of students be $n\geq 5$ . Then the sum of their scores is at least $5\cdot 100 + (n-5)\cdot 60$ . At the same time, we need to achieve the mean $76$ , which is equivalent to achieving the sum $76n$
Hence we get a necessary condition on $n$ : we must have $5\cdot 100 + (n-5)\cdot 60 \leq 76n$ .
This can be simplified to $200 \leq 16n$ . The smallest integer $n$ for which this is true is $n=13$
To finish our solution, we now need to find one way how $13$ students could have scored on the test. We have $13\cdot 76 = 988$ points to divide among them. The five $100$ s make $500$ , hence we must divide the remaining $488$ points among the other $8$ students. This can be done e.g. by giving $61$ points to each of them.
Hence the smallest possible number of students is $\boxed{13}$ | D | 13 |
2c64367b29bd43e71935f946ad571d68 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_12 | In the sequence $2001$ $2002$ $2003$ $\ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$ . What is the $2004^\textrm{th}$ term in this sequence?
$\mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007$ | We already know that $a_1=2001$ $a_2=2002$ $a_3=2003$ , and $a_4=2000$ . Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$ $a_6=2003+2000-2005=1998$ $a_7=2000+2005-1998=2007$ , and so on.
We can now discover the following pattern: $a_{2k+1} = 2001+2k$ and $a_{2k}=2004-2k$ . This is easily proved by induction. It follows that $a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}$ | null | 0 |
2c64367b29bd43e71935f946ad571d68 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_12 | In the sequence $2001$ $2002$ $2003$ $\ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$ . What is the $2004^\textrm{th}$ term in this sequence?
$\mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007$ | Note that the recurrence $a_n+a_{n+1}-a_{n+2}~=~a_{n+3}$ can be rewritten as $a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}$
Hence we get that $a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= \cdots$ and also $a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= \cdots$
From the values given in the problem statement we see that $a_3=a_1+2$
From $a_1+a_2 = a_3+a_4$ we get that $a_4=a_2-2$
From $a_2+a_3 = a_4+a_5$ we get that $a_5=a_3+2$
Following this pattern, we get $a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}$ | null | 0 |
4d59c38a972d73f0de732571d4344ac1 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_13 | If $f(x) = ax+b$ and $f^{-1}(x) = bx+a$ with $a$ and $b$ real, what is the value of $a+b$
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 1 \qquad\mathrm{(E)}\ 2$ | Since $f(f^{-1}(x))=x$ , it follows that $a(bx+a)+b=x$ , which implies $abx + a^2 +b = x$ . This equation holds for all values of $x$ only if $ab=1$ and $a^2+b=0$
Then $b = -a^2$ . Substituting into the equation $ab = 1$ , we get $-a^3 = 1$ . Then $a = -1$ , so $b = -1$ , and \[f(x)=-x-1.\] Likewise \[f^{-1}(x)=-x-1.\] These are inverses to one another since \[f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.\] \[f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.\] Therefore $a+b=\boxed{2}$ | null | 2 |
275372ab0953871e9106f8c082b473fd | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_15 | The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
$\mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45$ | If Jack's current age is $\overline{ab}=10a+b$ , then Bill's current age is $\overline{ba}=10b+a$
In five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$
We are given that $10a+b+5=2(10b+a+5)$
Thus $8a=19b+5 \Rightarrow a=\dfrac{19b+5}{8}$
For $b=1$ we get $a=3$ . For $b=2$ and $b=3$ the value $\frac{19b+5}8$ is not an integer, and for $b\geq 4$ $a$ is more than $9$ . Thus the only solution is $(a,b)=(3,1)$ , and the difference in ages is $31-13=\boxed{18}$ | B | 18 |
275372ab0953871e9106f8c082b473fd | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_15 | The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
$\mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45$ | Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.
The age difference is $(10a+b)-(10b+a)=9(a-b)$ , hence it is a multiple of $9$ . Thus Bill's current age modulo $9$ must be $4$
Thus Bill's age is in the set $\{13,22,31,40,49,58,67,76,85,94\}$
As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options $\{13,49,58,67\}$
Checking each of them, we see that only $13$ works, and gives the solution $31-13=\boxed{18}$ | B | 18 |
7b9dbc776c11258cb7814874710c80f4 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_19 | A truncated cone has horizontal bases with radii $18$ and $2$ . A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?
$\mathrm{(A)}\ 6 \qquad\mathrm{(B)}\ 4\sqrt{5} \qquad\mathrm{(C)}\ 9 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 6\sqrt{3}$ | Consider a trapezoid (label it $ABCD$ as follows) cross-section of the truncate cone along a diameter of the bases:
Above, $E,F,$ and $G$ are points of tangency . By the Two Tangent Theorem, $BF = BE = 18$ and $CF = CG = 2$ , so $BC = 20$ . We draw $H$ such that it is the foot of the altitude $\overline{HD}$ to $\overline{AB}$
By the Pythagorean Theorem \[r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12\] Therefore, the answer is $\boxed{6}.$ | A | 6 |
7b9dbc776c11258cb7814874710c80f4 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_19 | A truncated cone has horizontal bases with radii $18$ and $2$ . A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?
$\mathrm{(A)}\ 6 \qquad\mathrm{(B)}\ 4\sqrt{5} \qquad\mathrm{(C)}\ 9 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 6\sqrt{3}$ | Create a trapezoid with inscribed circle $O$ exactly like in Solution #1, and extend lines $\overline{AD}$ and $\overline{BC}$ from the solution above and label the point at where they meet $H$ . Because $\frac{\overline{GC}}{\overline{BE}}$ $\frac{1}{9}$ $\frac{\overline{HG}}{\overline{HE}}$ $\frac{1}{9}$ . Let $\overline{HG} = x$ and $\overline{GE} = 8x$
Because these are radii, $\overline{GO} = \overline{OE} = \overline{OF} = 4x$ $\overline{OF} \perp \overline{BH}$ so $\overline{OF}^2 + \overline{FH}^2 = \overline{OH}^2$ . Plugging in, we get $4x^2 + \overline{FH}^2 = 5x^2$ so $\overline{FH} = 3x$ .Triangles $OFH$ and $BEH$ are similar so $\frac{\overline{OF}}{\overline{BE}} = \frac{\overline{FH}}{\overline{EH}}$ which gives us $\frac{4x}{18} = \frac{3x}{9x}$ . Solving for x, we get \[x = 1.5\] and \[4x =6\] . Thus, the answer is $\boxed{6}$ | A | 6 |
2f61ee97b9447314ee0f6e5a175b8298 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_22 | The square
is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $g$
$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 136$ | All the unknown entries can be expressed in terms of $b$ .
Since $100e = beh = ceg = def$ , it follows that $h = \frac{100}{b}, g = \frac{100}{c}$ ,
and $f = \frac{100}{d}$ . Comparing rows $1$ and $3$ then gives $50bc = 2 \cdot \frac{100}{b} \cdot \frac{100}{c}$ ,
from which $c = \frac{20}{b}$ .
Comparing columns $1$ and $3$ gives $50d \cdot \frac{100}{c}= 2c \cdot \frac{100}{d}$ ,
from which $d = \frac{c}{5} = \frac{4}{b}$ .
Finally, $f = 25b, g = 5b$ , and $e = 10$ . All the entries are positive integers
if and only if $b = 1, 2,$ or $4$ . The corresponding values for $g$ are $5, 10,$ and $20$ , and their sum is $\boxed{35}$ | C | 35 |
2f61ee97b9447314ee0f6e5a175b8298 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_22 | The square
is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $g$
$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 136$ | We know because this is a multiplicative magic square that each of the following are equal to each other: $100e=ceg=50dg=beh=2cf=50bc=def=2gh$
From this we know that $50dg=2hg$ , thus $h=25d$ .
Thus $beh=be(25d)$ and $be(25d)=100e$ . Thus $b=\frac{4}{d}$ From this we know that $50bc=(50)(\frac{4}{d})(c)=50dg$ . Thus $c=\frac{d^2g}{4}$ .
Now we know from the very beginning that $100e=ceg$ or $100=cg$ or $100=\frac{d^2g}{4}(g)$ or $\frac{d^2g^2}{4}$ . Rearranging the equation $100=\frac{d^2g^2}{4}$ we have $(d^2)(g^2)=400$ or $dg=20$ due to $d$ and $g$ both being positive. Now that $dg=20$ we find all pairs of positive integers that multiply to $20$ . There is $(d,g)= (20,1);(10,2);(5,4);(4,5);(2,10);(1,20)$ . Now we know that $b=\frac{4}{d}$ and b has to be a positive integer. Thus $d$ can only be $1$ $2$ , or $4$ . Thus $g$ can only be $20$ $10$ ,or $5$ . Thus sum of $20+10+5$ $35$ . The answer is $\boxed{35}$ | C | 35 |
990c75934e35979048a4cbb29f303c9c | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_23 | The polynomial $x^3 - 2004 x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $n$ are possible?
$\mathrm{(A)}\ 250,\!000 \qquad\mathrm{(B)}\ 250,\!250 \qquad\mathrm{(C)}\ 250,\!500 \qquad\mathrm{(D)}\ 250,\!750 \qquad\mathrm{(E)}\ 251,\!000$ | Let the roots be $r,s,r + s$ , and let $t = rs$ . Then
and by matching coefficients, $2(r + s) = 2004 \Longrightarrow r + s = 1002$ . Then our polynomial looks like \[x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0\] and we need the number of possible products $t = rs = r(1002 - r)$ . Because $m=t+1002^2$ is an integer, we also note that $t$ must be an integer.
Since $r > 0$ and $t > 0$ , it follows that $0 < t = r(1002-r) < 501^2 = 251001$ , with the endpoints not achievable because the roots must be distinct and positive. Because neither $r$ nor $1002-r$ can be an integer, there are $251,000 - 500 = \boxed{250,500}$ possible values of $n = -1002t$ | C | 250,500 |
990c75934e35979048a4cbb29f303c9c | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_23 | The polynomial $x^3 - 2004 x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $n$ are possible?
$\mathrm{(A)}\ 250,\!000 \qquad\mathrm{(B)}\ 250,\!250 \qquad\mathrm{(C)}\ 250,\!500 \qquad\mathrm{(D)}\ 250,\!750 \qquad\mathrm{(E)}\ 251,\!000$ | Letting the roots be $r$ $s$ , and $t$ , where $t = r+s$ , we see that by Vieta's Formula's, $2004 = r+s+t = t + t = 2t$ , and so $t = 1002$ . Therefore, $x-1002$ is a factor of $x^3 - 2004x^2 + mx + n$ . Letting $x = 0$ gives that $1002 \mid n$ because $x - 1002 \mid x^3 - 2004x^2 + mx + n$ . Letting $n = -1002a$ and noting that $x^3 - 2004x^2 + mx + n = (x-1002)(x^2 - bx + a)$ for some $b$ , we see that $b$ is the sum of the roots of $x^2 - bx + a$ $r$ and $s$ , and so $b = 1002$ . Now, we have that $x^2 - 1002x + a$ has roots $r$ and $s$ , and we wish to find the number of possible values of $a$ . By the quadratic formula, we see that \[\frac{1002 \pm \sqrt{1002^2 - 4a}}{2} = 501 \pm \sqrt{501^2 - a}\] are the two values of noninteger positive real numbers $r$ and $s$ , neither of which is equal to $1002$ . This information gives us that $0 < 501^2 - a < 501^2$ , and so since $501^2 - a$ is evidently not a square, we have $501^2 - 1 - 500 = 251,001 - 500 - 1 = \boxed{250,500}$ possible values of $n = 1002a$ | C | 250,500 |
398074003d8a1f9a0f08dcc621734d79 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_25 | Given that $2^{2004}$ is a $604$ digit number whose first digit is $1$ , how many elements of the set $S = \{2^0,2^1,2^2,\ldots ,2^{2003}\}$ have a first digit of $4$
$\mathrm{(A)}\ 194 \qquad \mathrm{(B)}\ 195 \qquad \mathrm{(C)}\ 196 \qquad \mathrm{(D)}\ 197 \qquad \mathrm{(E)}\ 198$ | Given $n$ digits, there must be exactly one power of $2$ with $n$ digits such that the first digit is $1$ . Thus $S$ contains $603$ elements with a first digit of $1$ . For each number in the form of $2^k$ such that its first digit is $1$ , then $2^{k+1}$ must either have a first digit of $2$ or $3$ , and $2^{k+2}$ must have a first digit of $4,5,6,7$ . Thus there are also $603$ numbers with first digit $\{2,3\}$ and $603$ numbers with first digit $\{4,5,6,7\}$ . By using complementary counting , there are $2004 - 3 \times 603 = 195$ elements of $S$ with a first digit of $\{8,9\}$ . Now, $2^k$ has a first digit of $\{8,9\}$ if and only if the first digit of $2^{k-1}$ is $4$ , so there are $\boxed{195}$ elements of $S$ with a first digit of $4$ | B | 195 |
2310026d60347e5e3a0144ef93919a9e | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_1 | What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$ | The first $2003$ even counting numbers are $2,4,6,...,4006$
The first $2003$ odd counting numbers are $1,3,5,...,4005$
Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$
$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$
$= 1+1+1+...+1 = \boxed{2003}$ | D | 2003 |
2310026d60347e5e3a0144ef93919a9e | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_1 | What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$ | Using the sum of an arithmetic progression formula, we can write this as $\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{2003}$ | D | 2003 |
2310026d60347e5e3a0144ef93919a9e | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_1 | What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$ | The formula for the sum of the first $n$ even numbers, is $S_E=n^{2}+n$ , (E standing for even).
Sum of first $n$ odd numbers, is $S_O=n^{2}$ , (O standing for odd).
Knowing this, plug $2003$ for $n$
$S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow$ $\boxed{2003}$ | D | 2003 |
2310026d60347e5e3a0144ef93919a9e | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_1 | What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$ | In the case that we don't know if $0$ is considered an even number, we note that it doesn't matter! The sum of odd numbers is $O=1+3+5+...+4005$ . And the sum of even numbers is either $E_1=0+2+4...+4004$ or $E_2=2+4+6+...+4006$ . When compared to the sum of odd numbers, we see that each of the $n$ th term in the series of even numbers differ by $1$ . For example, take series $O$ and $E_1$ . The first terms are $1$ and $0$ . Their difference is $|1-0|=1$ . Similarly, take take series $O$ and $E_2$ . The first terms are $1$ and $2$ . Their difference is $|1-2|=1$ . Since there are $2003$ terms in each set, the answer $\boxed{2003}$ | D | 2003 |
2310026d60347e5e3a0144ef93919a9e | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_1 | What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$ | We can pair each term of the sums - the first even number with the first odd number, the second with the second, and so forth. Then, there are 2003 pairs with a difference of 1 in each pair - 2-1 is 1, 4-3 is 1, 6-5 is 1, and so on. Then, the solution is $1 \cdot 2003$ , and the answer is $\boxed{2003}$ | D | 2003 |
746ddcc2fdf4bd802d7bcae41fb87c55 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_2 | Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $4 per pair and each T-shirt costs $5 more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is $2366, how many members are in the League?
$\mathrm{(A) \ } 77\qquad \mathrm{(B) \ } 91\qquad \mathrm{(C) \ } 143\qquad \mathrm{(D) \ } 182\qquad \mathrm{(E) \ } 286$ | Since T-shirts cost $5$ dollars more than a pair of socks, T-shirts cost $5+4=9$ dollars.
Since each member needs $2$ pairs of socks and $2$ T-shirts, the total cost for $1$ member is $2(4+9)=26$ dollars.
Since $2366$ dollars was the cost for the club, and $26$ was the cost per member, the number of members in the League is $2366\div 26=\boxed{91}$ | B | 91 |
0bb62a753335aa41cea883511f7543f3 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_3 | A solid box is $15$ cm by $10$ cm by $8$ cm. A new solid is formed by removing a cube $3$ cm on a side from each corner of this box. What percent of the original volume is removed?
$\mathrm{(A) \ } 4.5\%\qquad \mathrm{(B) \ } 9\%\qquad \mathrm{(C) \ } 12\%\qquad \mathrm{(D) \ } 18\%\qquad \mathrm{(E) \ } 24\%$ | The volume of the original box is $15\cdot10\cdot8=1200.$
The volume of each cube that is removed is $3\cdot3\cdot3=27.$
Since there are $8$ corners on the box, $8$ cubes are removed.
So the total volume removed is $8\cdot27=216$
Therefore, the desired percentage is $\frac{216}{1200}\cdot100 = \boxed{18}.$ | D | 18 |
b79f694a655150ff2c02f8884cb4d317 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_4 | It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?
$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 4.5$ | Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km.
Since she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\frac{40}{60}=\frac{2}{3}$ hours.
Therefore her average speed in km/hr is $\frac{2}{\frac{2}{3}}=\boxed{3}$ | A | 3 |
b79f694a655150ff2c02f8884cb4d317 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_4 | It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?
$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 4.5$ | The average speed of two speeds that travel the same distance is the harmonic mean of the speeds, or $\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}=\dfrac{2xy}{x+y}$ (for speeds $x$ and $y$ ). Mary's speed going to school is $2\,\text{km/hr}$ , and her speed coming back is $6\,\text{km/hr}$ . Plugging the numbers in, we get that the average speed is $\dfrac{2\times 6\times 2}{6+2}=\dfrac{24}{8}=\boxed{3}$ | A | 3 |
ff034a832a33a8c73a14a86bc330ee63 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_5 | The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$
$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$ | $AMC10+AMC12=123422$
$AMC00+AMC00=123400$
$AMC+AMC=1234$
$2\cdot AMC=1234$
$AMC=\frac{1234}{2}=617$
Since $A$ $M$ , and $C$ are digits, $A=6$ $M=1$ $C=7$
Therefore, $A+M+C = 6+1+7 = \boxed{14}$ | E | 14 |
ff034a832a33a8c73a14a86bc330ee63 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_5 | The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$
$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$ | We know that $AMC12$ is $2$ more than $AMC10$ . We set up $AMC10=x$ and $AMC12=x+2$ . We have $x+x+2=123422$ . Solving for $x$ , we get $x=61710$ . Therefore, the sum $A+M+C= \boxed{14}$ | E | 14 |
ff034a832a33a8c73a14a86bc330ee63 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_5 | The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$
$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$ | Consider the place values of the digits of $AMC10$ and $AMC12$
When we add $AMC10$ and $AMC12$ $C + C$ must result in a units digit of $4$ , meaning $C$ is either $2$ or $7$ . Since $M$ is odd, this means a ten was carried over to the next place value from $C + C$ , and thus $C = 7$ (as $7 + 7 = 14$ and the ten is carried over). Now, we know 3 is the units digit of $M + M + 1$ , so $M$ is either $1$ or $6$ . Again, we must look at the digit before $M$ , or $A$ $A$ is even, so $M$ must be less than $5$ , or else the ten would be carried over. Ergo, $M$ is $1$ . Nothing is carried over, so we have $A + A = 12$ , and $A = 6$ . Therefore, the sum of $A$ $M$ , and $C$ is $6 + 1 + 7 = \boxed{14}$ | E | 14 |
a96f0e383b0940adee69c346f646b968 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_6 | Define $x \heartsuit y$ to be $|x-y|$ for all real numbers $x$ and $y$ . Which of the following statements is not true?
$\mathrm{(A) \ } x \heartsuit y = y \heartsuit x$ for all $x$ and $y$
$\mathrm{(B) \ } 2(x \heartsuit y) = (2x) \heartsuit (2y)$ for all $x$ and $y$
$\mathrm{(C) \ } x \heartsuit 0 = x$ for all $x$
$\mathrm{(D) \ } x \heartsuit x = 0$ for all $x$
$\mathrm{(E) \ } x \heartsuit y > 0$ if $x \neq y$ | We start by looking at the answers. Examining statement C, we notice:
$x \heartsuit 0 = |x-0| = |x|$
$|x| \neq x$ when $x<0$ , but statement C says that it does for all $x$
Therefore the statement that is not true is $\boxed{0}$ | C | 0 |
faedc2419bb889f494bddd75322b5a7d | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_7 | How many non- congruent triangles with perimeter $7$ have integer side lengths?
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5$ | By the triangle inequality , no side may have a length greater than the semiperimeter, which is $\frac{1}{2}\cdot7=3.5$
Since all sides must be integers, the largest possible length of a side is $3$ . Therefore, all such triangles must have all sides of length $1$ $2$ , or $3$ . Since $2+2+2=6<7$ , at least one side must have a length of $3$ . Thus, the remaining two sides have a combined length of $7-3=4$ . So, the remaining sides must be either $3$ and $1$ or $2$ and $2$ . Therefore, the number of triangles is $\boxed{2}$ | B | 2 |
836d15a5c91819ef5f2610d041f90797 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_9 | A set $S$ of points in the $xy$ -plane is symmetric about the origin, both coordinate axes, and the line $y=x$ . If $(2,3)$ is in $S$ , what is the smallest number of points in $S$
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 16$ | If $(2,3)$ is in $S$ , then $(3,2)$ is also, and quickly we see that every point of the form $(\pm 2, \pm 3)$ or $(\pm 3, \pm 2)$ must be in $S$ . Now note that these $8$ points satisfy all of the symmetry conditions. Thus the answer is $\boxed{8}$ | D | 8 |
7cfe4ce075a6df4d99fca4f598e3b6f4 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_12 | Sally has five red cards numbered $1$ through $5$ and four blue cards numbered $3$ through $6$ . She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?
$\mathrm{(A) \ } 8\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } 12$ | Let $R_i$ and $B_j$ designate the red card numbered $i$ and the blue card numbered $j$ , respectively.
$B_5$ is the only blue card that $R_5$ evenly divides, so $R_5$ must be at one end of the stack and $B_5$ must be the card next to it.
$R_1$ is the only other red card that evenly divides $B_5$ , so $R_1$ must be the other card next to $B_5$
$B_4$ is the only blue card that $R_4$ evenly divides, so $R_4$ must be at one end of the stack and $B_4$ must be the card next to it.
$R_2$ is the only other red card that evenly divides $B_4$ , so $R_2$ must be the other card next to $B_4$
$R_2$ doesn't evenly divide $B_3$ , so $B_3$ must be next to $R_1$ $B_6$ must be next to $R_2$ , and $R_3$ must be in the middle.
This yields the following arrangement from top to bottom: $\{R_5,B_5,R_1,B_3,R_3,B_6,R_2,B_4,R_4\}$
Therefore, the sum of the numbers on the middle three cards is $3+3+6=\boxed{12}$ | E | 12 |
380451ea9c7157f1bfe54b094b8a92b9 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_13 | The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge -to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
2003amc10a10.gif
$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$ | 2003amc10a10solution.gif
Let the squares be labeled $A$ $B$ $C$ , and $D$
When the polygon is folded, the "right" edge of square $A$ becomes adjacent to the "bottom edge" of square $C$ , and the "bottom" edge of square $A$ becomes adjacent to the "bottom" edge of square $D$
So, any "new" square that is attatched to those edges will prevent the polygon from becoming a cube with one face missing.
Therefore, squares $1$ $2$ , and $3$ will prevent the polygon from becoming a cube with one face missing.
Squares $4$ $5$ $6$ $7$ $8$ , and $9$ will allow the polygon to become a cube with one face missing when folded.
Thus the answer is $\boxed{6}$ | E | 6 |
380451ea9c7157f1bfe54b094b8a92b9 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_13 | The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge -to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
2003amc10a10.gif
$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$ | Another way to think of it is that a cube missing one face has $5$ of its $6$ faces. Since the shape has $4$ faces already, we need another face. The only way to add another face is if the added square does not overlap any of the others. $1$ $2$ , and $3$ overlap, while squares $4$ to $9$ do not. The answer is $\boxed{6}$ | E | 6 |
380451ea9c7157f1bfe54b094b8a92b9 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_13 | The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge -to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
2003amc10a10.gif
$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$ | If you're good at visualizing, you can imagine each box and fold up the shape into a 3D shape. This solution is only recommended if you are either in a hurry or extremely skilled at visualizing. We find out that $4,5,6,7,8$ and $9$ work. Therefore, the answer is $\boxed{6}$ . ~Sophia866 | E | 6 |
4c4910e6af26acb47fb9f996256a039b | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_18 | Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$ . For how many values of $n$ is $q+r$ divisible by $11$
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$ | When a $5$ -digit number is divided by $100$ , the first $3$ digits become the quotient, $q$ , and the last $2$ digits become the remainder, $r$
Therefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive.
For each of the $9\cdot10\cdot10=900$ possible values of $q$ , there are at least $\left\lfloor \frac{100}{11} \right\rfloor = 9$ possible values of $r$ such that $q+r \equiv 0\pmod{11}$
Since there is $1$ "extra" possible value of $r$ that is congruent to $0\pmod{11}$ , each of the $\left\lfloor \frac{900}{11} \right\rfloor = 81$ values of $q$ that are congruent to $0\pmod{11}$ have $1$ more possible value of $r$ such that $q+r \equiv 0\pmod{11}$
Therefore, the number of possible values of $n$ such that $q+r \equiv 0\pmod{11}$ is $900\cdot9+81\cdot1=8181 \Rightarrow\boxed{8181}$ | B | 8181 |
4c4910e6af26acb47fb9f996256a039b | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_18 | Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$ . For how many values of $n$ is $q+r$ divisible by $11$
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$ | Notice that $q+r\equiv0\pmod{11}\Rightarrow100q+r\equiv0\pmod{11}$ . This means that any number whose quotient and remainder sum is divisible by 11 must also be divisible by 11. Therefore, there are $\frac{99990-10010}{11}+1=8181$ possible values. The answer is $\boxed{8181}$ | B | 8181 |
4c4910e6af26acb47fb9f996256a039b | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_18 | Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$ . For how many values of $n$ is $q+r$ divisible by $11$
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$ | Let $abcde$ be the five digits of $n$ . Then $q = abc$ and $r = de$ . By the divisibility rules of $11$ $q = a - b + c \pmod{11}$ and $r = -d + e \pmod{11}$ , so $q + r = a - b + c - d + e = abcde = n \pmod{11}$ . Thus, $n$ must be divisble by $11$ . There are $\frac{99990 - 10010}{11} + 1 = 8181$ five-digit multiples of $11$ , so the answer is $\boxed{8181}$ | B | 8181 |
fb77f528521f533ff3d87e44d43ad352 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_22 | Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ and each of its steps is either to the left or down, both equally likely. Which of the following is closest to the probability that the objects meet?
$\mathrm{(A)} \ 0.10 \qquad \mathrm{(B)} \ 0.15 \qquad \mathrm{(C)} \ 0.20 \qquad \mathrm{(D)} \ 0.25 \qquad \mathrm{(E)} \ 0.30 \qquad$ | We know that the sum of the vertical steps must be equal to $7$ . We also know that they must take $6$ steps each. Since moving vertically or horizontally is equally likely, we can write all the possible paths as a generating function:
\[P(x)=(x+1)^{12}\]
Where we need to extract the $x^5$ coefficient. By the binomial coefficient theorem, that term is $\binom{12}{5}=792$ paths. Since there are also $2^{12}$ paths, we have:
$\frac{792}{2^{12}}=\frac{792}{4096}\approx\frac{800}{4000}\approx\boxed{0.20}$ | C | 0.20 |
46e5e72c3aad60eb47e7a50e81b5499f | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_23 | How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!$
$\textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008$ | We want to find the number of perfect square factors in the product of all the factorials of numbers from $1 - 9$ . We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. This comes out to be equal to $2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$ . To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: $2^{15} \cdot 3^{6}\cdot 5^2 \cdot 7^1$ . To find the total number of possibilities, we add $1$ to each exponent and multiply them all together. This gives us $16 \cdot 7 \cdot 3 \cdot 2 = 672$ $\Rightarrow\boxed{672}$ | B | 672 |
46e5e72c3aad60eb47e7a50e81b5499f | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_23 | How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!$
$\textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008$ | (Explanation of 1)
Factorials up to 9 in product lead to prime factorization $2^{30}$ $*$ $3^{13}$ $*$ $5^5$ $*$ $7^3$ So the number of pairs possible is { $2^{0}$ $,$ $2^{2}$ $,$ ... $2^{30}$ }{ $3^{0}$ $,$ $3^{2}$ $,$ ... $3^{12}$ }{ $5^{0}$ $,$ $5^{2}$ $,$ ... $5^{4}$ }{ $7^{0}$ $,$ $7^{2}$
Which leads to resulting number of pairs = $16*7*3*2$ $=$ $672$ $\Rightarrow\boxed{672}$ | B | 672 |
1300857de7bb256e3aed5a0bffb2229e | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_24 | If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$
$\mathrm{(A)}\ -2 \qquad \mathrm{(B)}\ 0 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 3 \qquad \mathrm{(E)}\ 4$ | Using logarithmic rules, we see that
\[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\] \[=2-\left(\log_{a}b+\frac {1}{\log_{a}b}\right)\]
Since $a$ and $b$ are both greater than $1$ , using AM-GM gives that the term in parentheses must be at least $2$ , so the largest possible values is $2-2=0 \Rightarrow \boxed{0}.$ | B | 0 |
1300857de7bb256e3aed5a0bffb2229e | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_24 | If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$
$\mathrm{(A)}\ -2 \qquad \mathrm{(B)}\ 0 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 3 \qquad \mathrm{(E)}\ 4$ | Similar to the previous solution, we use our logarithmic rules and start off the same way. \[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\]
However, now, we use a different logarithmic rule stating that $\log_{a}b$ is simply equal to $\frac {\log_{10}b}{\log_{10}a}$ . With this, we can rewrite our previous equation to give us \[2-(\log_{a}b+\log_{b}a) = 2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right)\]
We can now cross multiply to get that \[2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right) = 2 - \left(\frac {2 \cdot \log_{10}b \cdot \log_{10}a}{\log_{10}b \cdot \log_{10}a}\right)\] Finally, we cancel to get $2-2=0 \Rightarrow \boxed{0}.$ | B | 0 |
1300857de7bb256e3aed5a0bffb2229e | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_24 | If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$
$\mathrm{(A)}\ -2 \qquad \mathrm{(B)}\ 0 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 3 \qquad \mathrm{(E)}\ 4$ | By the logarithmic rules, we have $2-(\log_a{b}+\frac{1}{\log_a{b}})$ .
Let $x=\log_a{b}$ . Thus, the expression becomes $2-(x+\frac{1}{x})$ . We want to find the maximum of the function. To do so, we will find its derivative and let it equal to 0. We get: $\frac{d}{dx}\big(2-(x+\frac{1}{x})\big)=\frac{d}{dx}\big(2-x-\frac{1}{x})=-1+x^{-2}=0 \implies \frac{1}{x^2}=1, x^2=1, x=\pm1.$ Since $a\geq b>1, x=-1$ is not a solution. Thus, $x=1$ . Substituting it into the original expression $2-(x+\frac{1}{x})$ , we get $2-(1+\frac{1}{1})=2-2=\boxed{0}$ | null | 0 |
8c35fcff4ef77b2c44dd40ec88c46cde | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_25 | Let $f(x)= \sqrt{ax^2+bx}$ . For how many real values of $a$ is there at least one positive value of $b$ for which the domain of $f$ and the range of $f$ are the same set
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} }$ | If $f(x)=y$ , then squaring both sides of the given equation and subtracting $ax^2$ and $bx$ yields $y^2-ax^2-bx=0$ . Completing the square, we get $(x+\frac{b}{2a})^2-\frac{y^2}{a}=\frac{b^2}{4a^2}$ where $y\geq 0$ . Divide out by $\frac{b^2}{4a^2}$ to put the equation in the standard form of an ellipse or hyperbola (depending on the sign of $a$ ) to get $\frac{(x+\frac{b}{2a})^2}{(\frac{b}{2a})^2}-\frac{y^2}{(\frac{b}{2\sqrt {\pm a}})^2}=1$
Before continuing, it is important to note that because $f(x)=\sqrt{ax^2+bx}=\sqrt{ax(x+\frac{b}{a})}$ $f(x)$ has roots 0 and $-\frac{b}{a}$ . Now, we can use the function we deduced to figure out some of its properties when:
$a>0$ : A semi-hyperbola above or on the x-axis. Therefore, no positive value of $a$ allows the domain and range to be the same set because the range will always be $[0, \infty)$ and the domain will always be undefined on some finite range between some value and zero.
$a=0$ : We must refer back to the original function; this results in a horizontal semiparabola in the first quadrant, which satisfies that the domain and range of the function are equal. Specifically, both sets are $[0, \infty )$ $a=0$ is the only case where this happens.
$a<0$ : A semi-ellipse in quadrant one. Since its roots are 0 and $-\frac{b}{a}$ , its domain must be $[0, -\frac{b}{a}]$ . To make its domain and range equal, the maximum value of the ellipse must then be $-\frac{b}{a}$ . But we have another expression for the maximum value of the ellipse, which is $0+\frac{b}{2\sqrt {\pm a}}$ . Setting these two expressions equal to each other will find us the final value of $a$ that satisfies the question.
$\frac{b}{-a}=\frac{b}{2\sqrt {\pm a}}$
$-a=2\sqrt {\pm a}$
$a^2=\pm 4a$
$a=0,\pm 4$
We already knew 0 was a solution from earlier, so -4 is our only new solution (we already ruled out any positive value of $a$ as a solution, so 4 does not work). Thus there are $\boxed{2}$ values of $a$ that make the domain and range of $f(x)$ the same set. | C | 2 |
0ca8a6e48e2e27d4181ae08a4e8d9a08 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_2 | Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs $$ $1$ more than a pink pill, and Al's pills cost a total of $\textdollar 546$ for the two weeks. How much does one green pill cost?
$\textbf{(A)}\ \textdollar 7 \qquad\textbf{(B) }\textdollar 14 \qquad\textbf{(C) }\textdollar 19\qquad\textbf{(D) }\textdollar 20\qquad\textbf{(E) }\textdollar 39$ | Because there are $14$ days in two weeks, Al spends $546/14 = 39$ dollars per day for the cost of a green pill and a pink pill. If the green pill costs $x$ dollars and the pink pill $x-1$ dollars, the sum of the two costs $2x-1$ should equal $39$ dollars. Then the cost of the green pill $x$ is $\boxed{20}$ | D | 20 |
01758c6bbac4979dc5ed971135c49ad5 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_3 | Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost $$ $1$ each, begonias $$ $1.50$ each, cannas $$ $2$ each, dahlias $$ $2.50$ each, and Easter lilies $$ $3$ each. What is the least possible cost, in dollars, for her garden?
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$\textbf{(A) } 108 \qquad\textbf{(B) } 115 \qquad\textbf{(C) } 132 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 156$ | The areas of the five regions from greatest to least are $21,20,15,6$ and $4$
If we want to minimize the cost, we want to maximize the area of the cheapest flower and minimize the area of the most expensive flower. Doing this, the cost is $1\cdot21+1.50\cdot20+2\cdot15+2.50\cdot6+3\cdot4$ , which simplifies to $$ $108$ .
Therefore the answer is $\boxed{108}$ | A | 108 |
a7daf3b4a9cfb62d052477cd13633c61 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_4 | Moe uses a mower to cut his rectangular $90$ -foot by $150$ -foot lawn. The swath he cuts is $28$ inches wide, but he overlaps each cut by $4$ inches to make sure that no grass is missed. He walks at the rate of $5000$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow the lawn?
$\textbf{(A) } 0.75 \qquad\textbf{(B) } 0.8 \qquad\textbf{(C) } 1.35 \qquad\textbf{(D) } 1.5 \qquad\textbf{(E) } 3$ | Since the swath Moe actually mows is $24$ inches, or $2$ feet wide, he mows $10000$ square feet in one hour. His lawn has an area of $13500$ , so it will take Moe $1.35$ hours to finish mowing the lawn. Thus the answer is $\boxed{1.35}$ | C | 1.35 |
a7daf3b4a9cfb62d052477cd13633c61 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_4 | Moe uses a mower to cut his rectangular $90$ -foot by $150$ -foot lawn. The swath he cuts is $28$ inches wide, but he overlaps each cut by $4$ inches to make sure that no grass is missed. He walks at the rate of $5000$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow the lawn?
$\textbf{(A) } 0.75 \qquad\textbf{(B) } 0.8 \qquad\textbf{(C) } 1.35 \qquad\textbf{(D) } 1.5 \qquad\textbf{(E) } 3$ | Let's assume that the swath moves back and forth; parallel to the $90$ feet side. Thus, the length of one strip is $90$ feet. Now we need to find out how many strips there are. In reality, the swath Moe mows is $24$ inches wide, which can be easily translated into $2$ feet. $\frac{150}{2}$ is the number of strips Moe needs to mow, which is equal to $75$ .
Therefore, the total number of feet Moe mows is $75\times90$ . Since Moe's mowing rate is $5000$ feet per hour, $\frac{75\times90}{5000}$ is the number of hours it takes him to do his job. Using basic calculations, we compute the answer. $\boxed{1.35}$ | C | 1.35 |
8df11adead18a6429903cfb77f58b993 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_5 | Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $4:3$ . The horizontal length of a " $27$ -inch" television screen is closest, in inches, to which of the following?
$\textbf{(A) } 20 \qquad\textbf{(B) } 20.5 \qquad\textbf{(C) } 21 \qquad\textbf{(D) } 21.5 \qquad\textbf{(E) } 22$ | If you divide the television screen into two right triangles, the legs are in the ratio of $4 : 3$ , and we can let one leg be $4x$ and the other be $3x$ . Then we can use the Pythagorean Theorem.
\begin{align*}(4x)^2+(3x)^2&=27^2\\ 16x^2+9x^2&=729\\ 25x^2&=729\\ x^2&=\frac{729}{25}\\ x&=\frac{27}{5}\\ x&=5.4\end{align*}
The horizontal length is $5.4\times4=21.6$ , which is closest to $\boxed{21.5}$ | D | 21.5 |
8df11adead18a6429903cfb77f58b993 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_5 | Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $4:3$ . The horizontal length of a " $27$ -inch" television screen is closest, in inches, to which of the following?
$\textbf{(A) } 20 \qquad\textbf{(B) } 20.5 \qquad\textbf{(C) } 21 \qquad\textbf{(D) } 21.5 \qquad\textbf{(E) } 22$ | One can realize that the diagonal, vertical, and horizontal lengths all make up a $3,4,5$ triangle. Therefore, the horizontal length, being the $4$ in the $4 : 3$ ratio, is simply $\frac{4}{5}$ times the hypotenuse. $\frac{4}{5}\cdot27=21.6 \approx \boxed{21.5}$ | D | 21.5 |
a5a8575f3dcbbd1f65efff33b50a99a0 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_7 | Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, all nickels,dimes, and quarters, whose total value is $8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?
$\text {(A) } 0 \qquad \text {(B) } 13 \qquad \text {(C) } 37 \qquad \text {(D) } 64 \qquad \text {(E) } 83$ | Where $a,b,c$ is the number of nickels, dimes, and quarters, respectively, we can set up two equations:
\[(1)\ 5a+10b+25c=835\ \ \ \ (2)\ a+b+c=100\]
Eliminate $a$ by subtracting $(2)$ from $(1)/5$ to get $b+4c=67$ . Of the integer solutions $(b,c)$ to this equation, the number of dimes $b$ is least in $(3,16)$ and greatest in $(67,0)$ , yielding a difference of $67-3=\boxed{64}$ | D | 64 |
508dffca8aa46ac9dec59109e6bca392 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_8 | Let $\clubsuit(x)$ denote the sum of the digits of the positive integer $x$ . For example, $\clubsuit(8)=8$ and $\clubsuit(123)=1+2+3=6$ . For how many two-digit values of $x$ is $\clubsuit(\clubsuit(x))=3$
$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 9 \qquad\textbf{(E) } 10$ | Let $y=\clubsuit (x)$ . Since $x \leq 99$ , we have $y \leq 18$ . Thus if $\clubsuit (y)=3$ , then $y=3$ or $y=12$ . The 3 values of $x$ for which $\clubsuit (x)=3$ are 12, 21, and 30, and the 7 values of $x$ for which $\clubsuit (x)=12$ are 39, 48, 57, 66, 75, 84, and 93. There are $\boxed{10}$ values in all. | E | 10 |
e6d8b9804b11898aa89c2013ed1682bc | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_10 | Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way?
$\text {(A) } 1 \qquad \text {(B) } 2 \qquad \text {(C) } 3 \qquad \text {(D) } 4 \qquad \text {(E) } 5$ | Place the first triangle. Now, we can place the second triangle either adjacent to the first, or with one side between them, for a total of $\boxed{2}$ | B | 2 |
e6d8b9804b11898aa89c2013ed1682bc | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_10 | Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way?
$\text {(A) } 1 \qquad \text {(B) } 2 \qquad \text {(C) } 3 \qquad \text {(D) } 4 \qquad \text {(E) } 5$ | Take ${5 \choose 2}$ to realize there are 10 ways to choose 2 different triangles. Then divide by 5 for each vertex of a pentagon to get $\boxed{2}$ | B | 2 |
b29fbe18dd7e13b841bb6c816a281bff | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_12 | What is the largest integer that is a divisor of
\[(n+1)(n+3)(n+5)(n+7)(n+9)\]
for all positive even integers $n$
$\text {(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165$ | For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is $3 \cdot 5 = 15$ , so ${\boxed{15}$ is the correct answer. | D | 15 |
b29fbe18dd7e13b841bb6c816a281bff | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_12 | What is the largest integer that is a divisor of
\[(n+1)(n+3)(n+5)(n+7)(n+9)\]
for all positive even integers $n$
$\text {(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165$ | We'll just test all the answer choices.
Note that for any 3 consecutive odd integers, there will be exactly one multiple of $3.$
Let's list all possibilities of 3 consecutive odd integers. (multiple of 3, not multiple of 3, not multiple of 3), (not multiple of 3, multiple of 3, not multiple of 3) and (not multiple of 3, not multiple of 3, multiple of 3)
To support this further, list the first few consecutive lists of 3 consecutive odd integers.
We have $(1, 3, 5), (3, 5, 7), (5, 7, 9), (7, 9, 11), (11, 13, 15), \ldots$
So if the first two consecutive odd numbers aren't multiples of three, the last one must be a multiple of three.
Therefore for five consecutive odd integers, there must be at least one multiple of three.
In the same fashion as we did above, note that for any 5 consecutive integers, there will also be exactly one multiple of $5.$
Therefore, for any 5 consecutive odd integers, there must be exactly one multiple of five.
We can skip 7 since none of the answer choices are a multiple of 7.
Now we try $11.$ 11 doesn't work since as we see the first set of 5 consecutive odd integers doesn't fit, namely $(1, 3, 5, 7, 9).$
Since any 5 consecutive integers is divisible both by $3$ and $5$ , it also must be divisible by ${\boxed{15}$ and no higher since we saw that $11$ does not work and that there is no answer choice that is multiple of $7$ | D | 15 |
462512949a1f7a7d55e062d626b656f5 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_17 | If $\log (xy^3) = 1$ and $\log (x^2y) = 1$ , what is $\log (xy)$
$\mathrm{(A)}\ -\frac 12 \qquad\mathrm{(B)}\ 0 \qquad\mathrm{(C)}\ \frac 12 \qquad\mathrm{(D)}\ \frac 35 \qquad\mathrm{(E)}\ 1$ | We rewrite the logarithms in the problem. \[\log(x) + 3\log(y) = 1\] \[2\log(x) + \log(y) = 1\] \[\log(x) + \log(y) = c\] where $c$ is the desired quantity. Set $u = \log(x)$ and $v = \log(y)$ . Then we have that \[u + 3y = 1 \textbf{(1)}\] \[2u + y = 1 \textbf{(2)}\] \[u + v = c\] . Notice that \[2 \cdot \textbf{(2)} + \textbf{(1)} = 5u + 5v = 3 \implies u + v = \frac{3}{5} \implies c = \boxed{35}\] | D | 35 |
738eb5165f54d80bb2bc8dedb353daaa | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_18 | Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$ | Substitute $a^cb^d$ into $x$ . We then have $7(a^{5c}b^{5d}) = 11y^{13}$ . Divide both sides by $7$ , and it follows that:
\[(a^{5c}b^{5d}) = \frac{11y^{13}}{7}.\]
Note that because $11$ and $7$ are prime, the minimum value of $x$ must involve factors of $7$ and $11$ only. Thus, we try to look for the lowest power $p$ of $11$ such that $13p + 1 \equiv 0 \pmod{5}$ , so that we can take $11^{13p + 1}$ to the fifth root. Similarly, we want to look for the lowest power $n$ of $7$ such that $13n - 1 \equiv 0 \pmod{5}$ . Again, this allows us to take the fifth root of $7^{13n - 1}$ . Obviously, we want to add $1$ to $13p$ and subtract $1$ from $13n$ because $11^{13p}$ and $7^{13n}$ are multiplied by $11$ and divided by $7$ , respectively. With these conditions satisfied, we can simply multiply $11^{p}$ and $7^{n}$ and substitute this quantity into $y$ to attain our answer.
We can simply look for suitable values for $p$ and $n$ . We find that the lowest $p$ , in this case, would be $3$ because $13(3) + 1 \equiv 0 \pmod{5}$ . Moreover, the lowest $q$ should be $2$ because $13(2) - 1 \equiv 0 \pmod{5}$ . Hence, we can substitute the quantity $11^{3} \cdot 7^{2}$ into $y$ . Doing so gets us:
\[(a^{5c}b^{5d}) = \frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}.\]
Taking the fifth root of both sides, we are left with $a^cb^d = 11^{8} \cdot 7^{5}$ $a + b + c + d = 11 + 7 + 8 + 5 = \boxed{31}$ | B | 31 |
738eb5165f54d80bb2bc8dedb353daaa | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_18 | Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$ | A simpler way to tackle this problem without all that modding is to keep the equation as:
\[7*a^{5b}c^{5d} = 11y^{13}\]
As stated above, $a$ and $c$ must be the factors 7 and 11 in order to keep $x$ at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with:
\[7^{5b+1}11^{5d-1}=y^{13}\]
The above equation means that $y$ must also contain only the factors 7 and 11 (again, in order to keep $x$ at a minimum), so we end up with:
\[7^{5b+1}11^{5d-1}=7^{13h}11^{13j}\]
$h$ and $j$ are arbitrary variables placed in order to show that $y$ could have more than just one 7 or one 11 as factors)
Since 7 and 11 are prime, we know that $5b+1=13h$ and $5d-1=13j$ . The smallest positive combinations that would work are $b=5,h=2$ and $d=8,j=3$ . Therefore, $a+b+c+d=7+5+11+8=31$ $\boxed{31}$ is correct. | B | 31 |
738eb5165f54d80bb2bc8dedb353daaa | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_18 | Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$ | Another way to solve this problem solve for x. First, we can divide both sides by 7 to get:
\[x^5 = \frac{11y^{13}}{7}\]
Next, we take the fifth root on both sides, which gets us:
\[x = \sqrt[5]{\frac{11y^{13}}{7}}\]
\[x = y^2 \cdot \sqrt[5]{\frac{11y^{3}}{7}}\]
Since we know x is a positive integer that we are trying to minimize, we can let y equal the smallest number that will make x an integer. In this case, we let $y = 11^3 \cdot 7^2$ (Make sure you see why this makes x the smallest integer possible!), which when plugged in, results in:
\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{\frac{11}{7}\cdot (11^3 \cdot 7^2)^3}\]
\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{\frac{11}{7}\cdot 11^9 \cdot 7^6}\]
\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{11^{10} \cdot 7^5}\]
\[x = (11^3 \cdot 7^2)^2 \cdot 11^2 \cdot 7^1\]
\[x = 7^5 \cdot 11^ 8\]
This gets us $a^cb^d = 7^5\cdot 11^8$ , so $a + b + c + d = 7 + 5 + 11 + 8 = \boxed{31}$ ~lucaswujc, $\LaTeX$ help from Technodoggo | B | 31 |
738eb5165f54d80bb2bc8dedb353daaa | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_18 | Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$ | According to the problem, we have that $x^5$ and $y^{13}$ must be a multiple of both $7$ and $11$ . Thus, in their prime factorisation, there must be $7$ and $11$ . Thus, we have $a=7$ and $b=11$ . Now, let $x=7^c11^d\implies x^5=7^{5c}11^{5d}\implies7x^5=7^{5c+1}11^{5d}$ .
We can now divide both sides by 11 in our original equation $7x^5=11y^{13}$ to get $y^{13}=7^{5c+1}11^{5d-1}$ . As we are only considering integers, we have \[5c+1\equiv0~(\text{mod}~13)\implies5c\equiv12~(\text{mod}~13)\] and \[5d-1\equiv0~(\text{mod}~13)\implies5d\equiv1~(\text{mod}~13).\]
We can apply brute force to solve for $c$ and $d$ as the numbers aren't big. For the first congruence, we find that $25$ is the smallest number that satisfies, thus $c=5$ . For the second congruence, we find that $40$ is the smallest number that satisfies, thus $d=8$ . Summarising, we get $a+b+c+d=7+11+5+8=\boxed{31}$ | D | 31 |
c9764c3b77b5551032c0c0c690fe2115 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_20 | Part of the graph of $f(x) = ax^3 + bx^2 + cx + d$ is shown. What is $b$
2003 12B AMC-20.png
$\mathrm{(A)}\ -4 \qquad\mathrm{(B)}\ -2 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 2 \qquad\mathrm{(E)}\ 4$ | The roots of this equation are $-1, 1, \text{ and } x$ , letting $x$ be the root not shown in the graph. By Vieta, we know that $-1+1+x=x=-\frac{b}{a}$ and $-1\cdot 1\cdot x=-x=-\frac{d}{a}$ . Therefore, $x=\frac{d}{a}$ . Setting the two equations for $x$ equal to each other, $\frac{d}{a}=-\frac{b}{a}$ . We know that the y-intercept of the polynomial is $d$ , so $d=2$ . Plugging in for $d$ $\frac{2}{a}=-\frac{b}{a}$
Therefore, $b=-2 \Rightarrow \boxed{2}$ | B | 2 |
ccd43dfcc52e23c98877f8747815753b | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_21 | An object moves $8$ cm in a straight line from $A$ to $B$ , turns at an angle $\alpha$ , measured in radians and chosen at random from the interval $(0,\pi)$ , and moves $5$ cm in a straight line to $C$ . What is the probability that $AC < 7$
$\mathrm{(A)}\ \frac{1}{6} \qquad\mathrm{(B)}\ \frac{1}{5} \qquad\mathrm{(C)}\ \frac{1}{4} \qquad\mathrm{(D)}\ \frac{1}{3} \qquad\mathrm{(E)}\ \frac{1}{2}$ | $WLOG$ , let the object turn clockwise.
Let $B = (0, 0)$ $A = (0, -8)$
Note that the possible points of $C$ create a semi-circle of radius $5$ and center $B$ . The area where $AC < 7$ is enclosed by a circle of radius $7$ and center $A$ . The probability that $AC < 7$ is $\frac{\angle ABO}{180 ^\circ}$
The function of $\odot B$ is $x^2 + y^2 = 25$ , the function of $\odot A$ is $x^2 + (y+8)^2 = 49$
$O$ is the point that satisfies the system of equations: $\begin{cases} x^2 + y^2 = 25 \\ x^2 + (y+8)^2 = 49 \end{cases}$
$x^2 + (y+8)^2 - x^2 - y^2 = 49 - 25$ $64 + 16y =24$ $y = - \frac52$ $x = \frac{5 \sqrt{3}}{2}$ $O = (\frac{5 \sqrt{3}}{2}, - \frac52)$
Note that $\triangle BDO$ is a $30-60-90$ triangle, as $BO = 5$ $BD = \frac{5 \sqrt{3}}{2}$ $DO = \frac52$ . As a result $\angle CBO = 30 ^\circ$ $\angle ABO = 60 ^\circ$
Therefore the probability that $AC < 7$ is $\frac{\angle ABO}{180 ^\circ} = \frac{60 ^\circ}{180 ^\circ} = \boxed{13}$ | D | 13 |
391103ef657da99f744ce2ffbf258b53 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_22 | Let $ABCD$ be a rhombus with $AC = 16$ and $BD = 30$ . Let $N$ be a point on $\overline{AB}$ , and let $P$ and $Q$ be the feet of the perpendiculars from $N$ to $\overline{AC}$ and $\overline{BD}$ , respectively. Which of the following is closest to the minimum possible value of $PQ$
$\mathrm{(A)}\ 6.5 \qquad\mathrm{(B)}\ 6.75 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 7.25 \qquad\mathrm{(E)}\ 7.5$ | Let the intersection of $\overline{AC}$ and $\overline{BD}$ be $E$ . Since $ABCD$ is a rhombus, we have $\overline{AC} \perp \overline{BD}$ and $AE = CE = \dfrac{AC}{2} = 8$ . Since $\overline{NQ} \perp \overline{BD}$ , we have $\overline{NQ} \parallel \overline{AC}$ , so $\triangle{BNQ} \sim \triangle{BAE} \sim \triangle{NAP}$ . Therefore, \[\dfrac{NP}{AP} = \dfrac{NP}{8 - NQ} = \dfrac{BE}{AE} = \dfrac{15}{8} \Rightarrow NP = 15 - \dfrac{15}{8} NQ.\] By Pythagorean Theorem, \[PQ^2 = NQ^2 + NP^2 = NQ^2 + \left(15 - \dfrac{15}{8} NQ \right)^2 = \dfrac{289}{64} NQ^2 - \dfrac{225}{4} NQ + 225.\] The minimum value of $PQ^2$ would give the minimum value of $PQ$ , so we take the derivative (or use vertex form) to find that the minimum occurs when $NQ = \dfrac{225 \cdot 8}{289}$ which gives $PQ^2 = \dfrac{225 \cdot 64}{289}$ . Hence, the minimum value of $PQ$ is $\sqrt{\dfrac{225 \cdot 64}{289}} = \dfrac{120}{17}$ , which is closest to $7 \Rightarrow \boxed{7}$ | C | 7 |
391103ef657da99f744ce2ffbf258b53 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_22 | Let $ABCD$ be a rhombus with $AC = 16$ and $BD = 30$ . Let $N$ be a point on $\overline{AB}$ , and let $P$ and $Q$ be the feet of the perpendiculars from $N$ to $\overline{AC}$ and $\overline{BD}$ , respectively. Which of the following is closest to the minimum possible value of $PQ$
$\mathrm{(A)}\ 6.5 \qquad\mathrm{(B)}\ 6.75 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 7.25 \qquad\mathrm{(E)}\ 7.5$ | Let the intersection $\overline{AC}$ and $\overline{BD}$ be $E.$ Let $\overline{PE}=y \implies \overline{AP}=8-y$ and $\overline{EQ}=x \implies \overline {QB}=15-x.$ Since $\triangle NQB \sim \triangle APN,$ we have $\frac{8-y}{x}=\frac{y}{15-x} \implies 120=15y+8x.$ We want to minimize $\overline{PQ}=\sqrt{x^2+y^2}.$ By Cauchy, $17^2(x^2+y^2)=(x^2+y^2)(8^2+15^2) \ge (8x+15y)^2=120^2 \implies \sqrt{x^2+y^2} \ge \frac{120}{17} \approx 7.$ So, choice $\boxed{7}$ is closest to the minimum. | C | 7 |
d068e227e879cd3127b319bc02deebac | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_23 | The number of $x$ -intercepts on the graph of $y=\sin(1/x)$ in the interval $(0.0001,0.001)$ is closest to
$\mathrm{(A)}\ 2900 \qquad\mathrm{(B)}\ 3000 \qquad\mathrm{(C)}\ 3100 \qquad\mathrm{(D)}\ 3200 \qquad\mathrm{(E)}\ 3300$ | The function $f(x) = \sin x$ has roots in the form of $\pi n$ for all integers $n$ . Therefore, we want $\frac{1}{x} = \pi n$ on $\frac{1}{10000} \le x \le \frac{1}{1000}$ , so $1000 \le \frac 1x = \pi n \le 10000$ . There are $\frac{10000-1000}{\pi} \approx \boxed{2900}$ solutions for $n$ on this interval. | A | 2900 |
d068e227e879cd3127b319bc02deebac | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_23 | The number of $x$ -intercepts on the graph of $y=\sin(1/x)$ in the interval $(0.0001,0.001)$ is closest to
$\mathrm{(A)}\ 2900 \qquad\mathrm{(B)}\ 3000 \qquad\mathrm{(C)}\ 3100 \qquad\mathrm{(D)}\ 3200 \qquad\mathrm{(E)}\ 3300$ | We know that $x$ belongs to the interval $(0.0001,0.001)$ for $\sin(1/x)$ . We see that when we plug in $x$ into $\sin(1/x)$ , the argument $(1/x)$ is always from the range $(1000, 10000)$ . Therefore, the problem simply asks for all the zeros of $\sin(x)$ with $x$ values between $(1000, 10000)$ . We know that the $x$ values of any sine graph is $\pi(n-1)$ so, we see that values of $n$ are any integer value from $320$ to $3184$ and therefore gives us an answer of approximately $2865$ which is answer $\boxed{2900}$ | A | 2900 |
d9ba9b1ca08bee47e026577ce5441eee | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_24 | Positive integers $a,b,$ and $c$ are chosen so that $a<b<c$ , and the system of equations
has exactly one solution. What is the minimum value of $c$
$\mathrm{(A)}\ 668 \qquad\mathrm{(B)}\ 669 \qquad\mathrm{(C)}\ 1002 \qquad\mathrm{(D)}\ 2003 \qquad\mathrm{(E)}\ 2004$ | Consider the graph of $f(x)=|x-a|+|x-b|+|x-c|$
When $x<a$ , the slope is $-3$
When $a<x<b$ , the slope is $-1$
When $b<x<c$ , the slope is $1$
When $c<x$ , the slope is $3$
Setting $x=b$ gives $y=|b-a|+|b-b|+|b-c|=c-a$ , so $(b,c-a)$ is a point on $f(x)$ . In fact, it is the minimum of $f(x)$ considering the slope of lines to the left and right of $(b,c-a)$ . Thus, graphing this will produce a figure that looks like a cup: [asy] import graph; size(100); draw((0,6)--(3,0)); xaxis(0,8.5); yaxis(0,10); real f(real x) { return -3(x-2)+5; } real f2(real x) { return -1(x-2)+5; } real f3(real x) { return 1(x-4)+3; } real f4(real x) { return 3(x-7)+6; } draw(graph(f,0,2)); draw(graph(f2,2,4)); draw(graph(f3,4,7)); draw(graph(f4,7,8.5)); draw((2,-0.25)--(2,0.25)); label("a",(2,0),N); draw((4,-0.25)--(4,0.25)); label("b",(4,0),N); draw((7,-0.25)--(7,0.25)); label("c",(7,0),N); dot((2,5)); label("P",(1.9,5.2),E); [/asy] From the graph, it is clear that $f(x)$ and $2x+y=2003$ have one intersection point if and only if they intersect at $x=a$ . Since the line where $a<x<b$ has slope $-1$ , the positive difference in $y$ -coordinates from $x=a$ to $x=b$ must be $b-a$ . Together with the fact that $(b,c-a)$ is on $f(x)$ , we see that $P=(a,c-a+b-a)$ . Since this point is on $x=a$ , the only intersection point with $2x+y=2003$ , we have $2 \cdot a+(b+c-2a)=2003 \implies b+c=2003$ . As $c>b$ , the smallest possible value of $c$ occurs when $b=1001$ and $c=1002$ . This is indeed a solution as $a=1000$ puts $P$ on $y=2003-2x$ , and thus the answer is $\boxed{1002}$ | C | 1002 |
d9ba9b1ca08bee47e026577ce5441eee | https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_24 | Positive integers $a,b,$ and $c$ are chosen so that $a<b<c$ , and the system of equations
has exactly one solution. What is the minimum value of $c$
$\mathrm{(A)}\ 668 \qquad\mathrm{(B)}\ 669 \qquad\mathrm{(C)}\ 1002 \qquad\mathrm{(D)}\ 2003 \qquad\mathrm{(E)}\ 2004$ | We just showed that whenever $c<1002$ , the system has at least two different solutions: one with $x\leq a$ and one with $x>a$
We also showed that for $c=1002$ there are some $a,b$ for which the system has exactly one solution.
Hence the optimal value of $c$ is $\boxed{1002}$ | C | 1002 |
2e24e4bb084a5b560cea7873bba4f9fd | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_2 | Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
$\textbf{(A) } 15\qquad \textbf{(B) } 34\qquad \textbf{(C) } 43\qquad \textbf{(D) } 51\qquad \textbf{(E) } 138$ | We work backwards; the number that Cindy started with is $3(43)+9=138$ . Now, the correct result is $\frac{138-3}{9}=\frac{135}{9}=15$ . Our answer is $\boxed{15}$ | A | 15 |
2e24e4bb084a5b560cea7873bba4f9fd | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_2 | Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
$\textbf{(A) } 15\qquad \textbf{(B) } 34\qquad \textbf{(C) } 43\qquad \textbf{(D) } 51\qquad \textbf{(E) } 138$ | Let the number be $x$ . We transform the problem into an equation: $\frac{x-9}{3}=43$ . Solve for $x$ gives us $x=138$ . Therefore, the correct result is $\frac{138-3}{9}=\frac{135}{9}=\boxed{15}$ | A | 15 |
76a119fb12551b324a1a993bc1c44b32 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_3 | According to the standard convention for exponentiation, \[2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536.\]
If the order in which the exponentiations are performed is changed, how many other values are possible?
$\textbf{(A) } 0\qquad \textbf{(B) } 1\qquad \textbf{(C) } 2\qquad \textbf{(D) } 3\qquad \textbf{(E) } 4$ | The best way to solve this problem is by simple brute force.
It is convenient to drop the usual way how exponentiation is denoted, and to write the formula as $2\uparrow 2\uparrow 2\uparrow 2$ , where $\uparrow$ denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:
We can note that $2\uparrow (2\uparrow 2) = (2\uparrow 2)\uparrow 2 =16$ . Therefore options 1 and 2 are equal, and options 3 and 4 are equal.
Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.
$((2\uparrow 2)\uparrow 2)\uparrow 2 = 16\uparrow 2 = 256$
$(2\uparrow 2)\uparrow (2\uparrow 2) = 4 \uparrow 4 = 256$
Thus the only other result is $256$ , and our answer is $\boxed{1}$ | B | 1 |
e0ac70db00646afc0cf27788525faaab | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_5 | Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
[asy] unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]
$\textbf{(A)}\ \pi \qquad \textbf{(B)}\ 1.5\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ 3\pi \qquad \textbf{(E)}\ 3.5\pi$ | The outer circle has radius $1+1+1=3$ , and thus area $9\pi$ . The little circles have area $\pi$ each; since there are 7, their total area is $7\pi$ . Thus, our answer is $9\pi-7\pi=\boxed{2}$ | C | 2 |
f475ad6696c0cf4d95091c274532a1bc | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_9 | Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)} 16$ | $0.8$ MB file can either be on its own disk, or share it with a $0.4$ MB. Clearly it is better to pick the second possibility. Thus we will have $3$ disks, each with one $0.8$ MB file and one $0.4$ MB file.
We are left with $12$ files of $0.7$ MB each, and $12$ files of $0.4$ MB each. Their total size is $12\cdot (0.7 + 0.4) = 13.2$ MB. The total capacity of $9$ disks is $9\cdot 1.44 = 12.96$ MB, hence we need at least $10$ more disks. And we can easily verify that $10$ disks are indeed enough: six of them will carry two $0.7$ MB files each, and four will carry three $0.4$ MB files each.
Thus our answer is $3+10 = \boxed{13}$ | B | 13 |
f475ad6696c0cf4d95091c274532a1bc | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_9 | Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)} 16$ | Similarly to Solution 1, we see that there must be $3$ disks to account for the $0.8$ MB file. Secondly, since there are $[30-(3+12)]-3 = 12$ files(for both 0.4 MB and 0.7 MB) left, it is easy to see that the optimal way to place the files would be $3$ files per disks for the 0.4 MB files and hence would require 4 disks.
We are left with $12$ files( $0.7$ MB), where the optimal number of files per disks is $2$ , so the optimal number of disks for this type of file would be $6$ disks. Therefore, the answer is $3+4+6=\boxed{13}$ | null | 13 |
9c230344cd6db6741bcf29a93a04b4f3 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_10 | Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
$\mathrm{(A) \ } \frac{1}{4}\qquad \mathrm{(B) \ } \frac13\qquad \mathrm{(C) \ } \frac38\qquad \mathrm{(D) \ } \frac25\qquad \mathrm{(E) \ } \frac12$ | We will simulate the process in steps.
In the beginning, we have:
In the first step we pour $4/2=2$ ounces of coffee from cup $1$ to cup $2$ , getting:
In the second step we pour $2/2=1$ ounce of coffee and $4/2=2$ ounces of cream from cup $2$ to cup $1$ , getting:
Hence at the end we have $3+2=5$ ounces of liquid in cup $1$ , and out of these $2$ ounces is cream. Thus the answer is $\boxed{25}$ | D | 25 |
9c230344cd6db6741bcf29a93a04b4f3 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_10 | Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
$\mathrm{(A) \ } \frac{1}{4}\qquad \mathrm{(B) \ } \frac13\qquad \mathrm{(C) \ } \frac38\qquad \mathrm{(D) \ } \frac25\qquad \mathrm{(E) \ } \frac12$ | Let's consider this in steps.
We have 4 ounces of coffee in the first cup.
We hace 4 ounces of cream in the second cup.
We take half of the coffee in the first cup(2 ounces), and add it to the second cup, yielding 6 ounces in total in the second cup(in a 1:2 ratio between coffee and cream, respecitvely).
We then take half of the second cup and pour it into the first cup. $6/2=3$ , so there is now 5 ounces in the first cup, 2 coffee and 3 the mixture.
Remember that the mixture is in a 1:2 ratio between coffee and cream. So, coffee has one ounce and cream has 2 ounces.
In total, there is 2 ounces in the 5 ounce first cup. Putting 2 over 5, we get the answer. Therefore, the answer is $\boxed{25}$ | D | 25 |
547db35217b50658b7b174e033bac77e | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_11 | Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58$ | Let the time he needs to get there in be $t$ and the distance he travels be $d$ . From the given equations, we know that $d=\left(t+\frac{1}{20}\right)40$ and $d=\left(t-\frac{1}{20}\right)60$ . Setting the two equal, we have $40t+2=60t-3$ and we find $t=\frac{1}{4}$ of an hour. Substituting t back in, we find $d=12$ . From $d=rt$ , we find that $r$ , our answer, is $\boxed{48}$ | B | 48 |
547db35217b50658b7b174e033bac77e | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_11 | Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58$ | Since either time he arrives at is $3$ minutes from the desired time, the answer is merely the harmonic mean of 40 and 60.
Substituting $t=\frac ds$ and dividing both sides by $d$ , we get $\frac 2s = \frac 1{40} + \frac 1{60}$ hence $s=\boxed{48}$ | B | 48 |
547db35217b50658b7b174e033bac77e | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_11 | Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58$ | Let x be equal to the total amount of distance he needs to cover. Let y be equal to the amount of time he would travel correctly.
Setting up a system of equations, $\frac x{40} -3 = y$ and $\frac x{60} +3 = y$
Solving, we get x = 720 and y = 15.
We divide x by y to get the average speed, $\frac {720}{15} = 48$ . Therefore, the answer is $\boxed{48}$ | B | 48 |
27cb4207e25399857eb24816ed863855 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_12 | Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$ | Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$ . It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$ . Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to Vieta's Formulas ).
We now have that the sum of the two roots is $63$ while the product is $k$ . Since both roots are primes, one must be $2$ , otherwise, the sum would be even. That means the other root is $61$ and the product must be $122$ . Hence, our answer is $\boxed{1}$ | B | 1 |
27cb4207e25399857eb24816ed863855 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_12 | Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$ | By Vieta's you have \[a+b=63\] \[ab=k\] Since we know odd + even = odd, we must have either $a$ or $b$ equal to $2$ and the other equal to $63-2=61.$ Both of these are prime so they satisfy the restraints. Thus there is $\boxed{1}$ solution. | B | 1 |
b429fecabaeee298672d15739f552584 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_13 | Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$ . What is $a+b$
$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }\sqrt 5 \qquad \text{(D) }\sqrt 6 \qquad \text{(E) }3$ | Each of the numbers $a$ and $b$ is a solution to $\left| x - \frac 1x \right| = 1$
Hence it is either a solution to $x - \frac 1x = 1$ , or to $\frac 1x - x = 1$ . Then it must be a solution either to $x^2 - x - 1 = 0$ , or to $x^2 + x - 1 = 0$
There are in total four such values of $x$ , namely $\frac{\pm 1 \pm \sqrt 5}2$
Out of these, two are positive: $\frac{-1+\sqrt 5}2$ and $\frac{1+\sqrt 5}2$ . We can easily check that both of them indeed have the required property, and their sum is $\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{5}$ | C | 5 |
3e783ea1c0ddad231f9842c2e782af79 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_14 | For all positive integers $n$ , let $f(n)=\log_{2002} n^2$ . Let $N=f(11)+f(13)+f(14)$ . Which of the following relations is true?
$\text{(A) }N<1 \qquad \text{(B) }N=1 \qquad \text{(C) }1<N<2 \qquad \text{(D) }N=2 \qquad \text{(E) }N>2$ | First, note that $2002 = 11 \cdot 13 \cdot 14$
Using the fact that for any base we have $\log a + \log b = \log ab$ , we get that $N = \log_{2002} (11^2 \cdot 13^2 \cdot 14^2) = \log_{2002} 2002^2 = \boxed{2}$ | D | 2 |
872e48f7d607d28a5ee88ab4bfc8852f | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_15 | The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$ | As the unique mode is $8$ , there are at least two $8$ s.
As the range is $8$ and one of the numbers is $8$ , the largest one can be at most $16$
If the largest one is $16$ , then the smallest one is $8$ , and thus the mean is strictly larger than $8$ , which is a contradiction.
If we have 2 8's we can add find the numbers 4, 6, 7, 8, 8, 9, 10, 12.
This is a possible solution but has not reached the maximum.
If we have 4 8's we can find the numbers 6, 6, 6, 8, 8, 8, 8, 14.
We can also see that they satisfy the need for the mode, median, and range to be 8. This means that the answer will be $\boxed{14}$ . ~By QWERTYUIOPASDFGHJKLZXCVBNM | D | 14 |