id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
4,701 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_22 | 3 | Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textb... | For the sake of simplicity, let $R$ be a $60 \times 60$ square and set the bottom-left point as the origin. Then, $R$ has vertices: \[(0,0), (60,0), (60,60), (0,60).\]
Now, let a point in the square have coordinates $(x, y).$
In order for the point to be $100-$ ray partitional, we must be able to make $100$ triangles w... | 320 |
4,702 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_25 | 1 | Triangle $ABC$ has $\angle BAC = 60^{\circ}$ $\angle CBA \leq 90^{\circ}$ $BC=1$ , and $AC \geq AB$ . Let $H$ $I$ , and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$ , respectively. Assume that the area of pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$
$\textbf{(A)}\ 60^{\circ} \... | Let $\angle CAB=A$ $\angle ABC=B$ $\angle BCA=C$ for convenience.
It's well-known that $\angle BOC=2A$ $\angle BIC=90+\frac{A}{2}$ , and $\angle BHC=180-A$ (verifiable by angle chasing). Then, as $A=60$ , it follows that $\angle BOC=\angle BIC=\angle BHC=120$ and consequently pentagon $BCOIH$ is cyclic. Observe that $B... | 80 |
4,703 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_2 | 1 | Josanna's test scores to date are $90, 80, 70, 60,$ and $85.$ Her goal is to raise her test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?
$\textbf{(A)}\ 80 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 85 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}... | Take the average of her current test scores, which is \[\frac{90+80+70+60+85}{5} = \frac{385}{5} = 77\]
This means that she wants her test average after the sixth test to be $80.$ Let $x$ be the score that Josanna receives on her sixth test. Thus, our equation is
\[\frac{90+80+70+60+85+x}{6} = 80\]
\[385+x = 480\]
\[x ... | 95 |
4,704 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_3 | 1 | LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid A dollars and Bernardo had paid B dollars, where $A < B.$ How many dollars mus... | The total amount of money that was spent during the trip was \[A + B\] So each person should pay \[\frac{A+B}{2}\] if they were to share the costs equally. Because LeRoy has already paid $A$ dollars of his part, he still has to pay \[\frac{A+B}{2} - A =\] \[= \boxed{2}\] | 2 |
4,705 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_4 | 1 | In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161.$ What is the correct value of the product of $a$ and $b$
$\textbf{(A)}\ 116 \qquad \textbf{(B)}\ 161 \qquad \textbf{(C)}\ 204 \qquad \textbf{(D)}\ 214 \qquad \textbf{(E)}\ 224$ | Taking the prime factorization of $161$ reveals that it is equal to $23*7.$ Therefore, the only ways to represent $161$ as a product of two positive integers is $161*1$ and $23*7.$ Because neither $161$ nor $1$ is a two-digit number, we know that $a$ and $b$ are $23$ and $7.$ Because $23$ is a two-digit number, we know... | 224 |
4,706 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_5 | 1 | Let $N$ be the second smallest positive integer that is divisible by every positive integer less than $7$ . What is the sum of the digits of $N$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$ | $N$ must be divisible by every positive integer less than $7$ , or $1, 2, 3, 4, 5,$ and $6$ . Each number that is divisible by each of these is a multiple of their least common multiple. $LCM(1,2,3,4,5,6)=60$ , so each number divisible by these is a multiple of $60$ . The smallest multiple of $60$ is $60$ , so the seco... | 3 |
4,707 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_6 | 1 | Two tangents to a circle are drawn from a point $A$ . The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$ . What is the degree measure of $\angle{BAC}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60$ | In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).
In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to... | 36 |
4,708 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_6 | 2 | Two tangents to a circle are drawn from a point $A$ . The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$ . What is the degree measure of $\angle{BAC}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60$ | Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is $\frac{3x-2x}2 = \frac {x} 2$ by the arc length formula. Also note that $3x + 2x = 360$ , which simplifies to $x= 72.$ Hence the angle formed by the tangents is equal to $\boxed{36}$ | 36 |
4,709 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_6 | 3 | Two tangents to a circle are drawn from a point $A$ . The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$ . What is the degree measure of $\angle{BAC}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60$ | Let the center of the circle be $O$ . The radii extending from $O$ to the points of tangency $B$ and $C$ are both perpendicular to lines $AB$ and $AC$ . Thus $\angle ABO = \angle ACO = 90^{\circ}$ , and quadrilateral $ABOC$ is cyclic, since the opposite angles add to $180^{\circ}$
Since the ratio of the arc lengths is ... | 36 |
4,710 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_8 | 1 | Keiko walks once around a track at the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters ... | To find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the difference in the time it takes her for each distance. We are given the difference in time, so all we need to find is the dif... | 3 |
4,711 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_10 | 1 | Rectangle $ABCD$ has $AB=6$ and $BC=3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD=\angle CMD$ . What is the degree measure of $\angle AMD$
$\textrm{(A)}\ 15 \qquad \textrm{(B)}\ 30 \qquad \textrm{(C)}\ 45 \qquad \textrm{(D)}\ 60 \qquad \textrm{(E)}\ 75$ | Since $AB \parallel CD$ $\angle AMD = \angle CDM$ , so $\angle AMD = \angle CMD = \angle CDM$ , so $\bigtriangleup CMD$ is isosceles, and hence $CM=CD=6$ . Therefore, $\angle BMC = 30^\circ$ . Therefore $\angle AMD=\boxed{75}$ | 75 |
4,712 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_11 | 1 | A frog located at $(x,y)$ , with both $x$ and $y$ integers, makes successive jumps of length $5$ and always lands on points with integer coordinates. Suppose that the frog starts at $(0,0)$ and ends at $(1,0)$ . What is the smallest possible number of jumps the frog makes?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad... | Since the frog always jumps in length $5$ and lands on a lattice point, the sum of its coordinates must change either by $5$ (by jumping parallel to the x- or y-axis), or by $3$ or $4$ (3-4-5 right triangle).
Because either $1$ $5$ , or $7$ is always the change of the sum of the coordinates, the sum of the coordinates ... | 3 |
4,713 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_13 | 1 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$ . What is the sum of the possible values of $w$
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93$ | Assume that $y-z=a, x-y=b, w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$ .
This means $a+b+c=9$ $a,b,c$ must be in the set ${1,3,4,5,6}$ .
The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$ $a+b$ , and $b+c$ then must be the remaining two numbers which are $4$ a... | 31 |
4,714 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_15 | 1 | How many positive two-digit integers are factors of $2^{24}-1$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$
~ pi_is_3.14 | Repeating difference of squares
$2^{24}-1=(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)$
$2^{24}-1=(2^{12}+1)\cdot65\cdot9\cdot7$
$2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7$
The sum of cubes formula gives us:
$2^{12}+1=(2^4+1)(2^8-2^4+1)$
$2^{12}+1 = 17\cdot241$
A quick check shows $241$ is prime. Thus, the only factors to be ... | 12 |
4,715 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_17 | 1 | Let $f(x) = 10^{10x}, g(x) = \log_{10}\left(\frac{x}{10}\right), h_1(x) = g(f(x))$ , and $h_n(x) = h_1(h_{n-1}(x))$ for integers $n \geq 2$ . What is the sum of the digits of $h_{2011}(1)$
$\textbf{(A)}\ 16081 \qquad \textbf{(B)}\ 16089 \qquad \textbf{(C)}\ 18089 \qquad \textbf{(D)}\ 18098 \qquad \textbf{(E)}\ 18099$ | $g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1$
$h_{1}(x)=g(f(x))\text{ = }g(10^{10x}=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}$
Proof by induction that $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$
For $n=1$ $h_{1}(x)=10x - 1$
Assume $h_{n}(x)=10^n x - (1 + 10 + 10^2 + ..... | 89 |
4,716 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_17 | 2 | Let $f(x) = 10^{10x}, g(x) = \log_{10}\left(\frac{x}{10}\right), h_1(x) = g(f(x))$ , and $h_n(x) = h_1(h_{n-1}(x))$ for integers $n \geq 2$ . What is the sum of the digits of $h_{2011}(1)$
$\textbf{(A)}\ 16081 \qquad \textbf{(B)}\ 16089 \qquad \textbf{(C)}\ 18089 \qquad \textbf{(D)}\ 18098 \qquad \textbf{(E)}\ 18099$ | As before, $h_1(x)=10x-1$ . Compute $h_1(x)$ $h_2(x)$ , and $h_3(x)$ to yield 9, 89, and 889. Notice how this trend will repeat this trend (multiply by 10, subtract 1, repeat). As such, $h_{2011}$ is just 2010 8's followed by a nine. $2010(8)+9=\boxed{16089}$ | 89 |
4,717 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_23 | 1 | A bug travels in the coordinate plane, moving only along the lines that are parallel to the $x$ -axis or $y$ -axis. Let $A = (-3, 2)$ and $B = (3, -2)$ . Consider all possible paths of the bug from $A$ to $B$ of length at most $20$ . How many points with integer coordinates lie on at least one of these paths?
$\textbf{... | We declare a point $(x, y)$ to make up for the extra steps that the bug has to move. If the point $(x, y)$ satisfies the property that $|x - 3| + |y + 2| + |x + 3| + |y - 2| \le 20$ , then it is in the desirable range because $|x - 3| + |y + 2|$ is the length of the shortest path from $(x,y)$ to $(3, -2)$ and $|x + 3| ... | 195 |
4,718 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_23 | 2 | A bug travels in the coordinate plane, moving only along the lines that are parallel to the $x$ -axis or $y$ -axis. Let $A = (-3, 2)$ and $B = (3, -2)$ . Consider all possible paths of the bug from $A$ to $B$ of length at most $20$ . How many points with integer coordinates lie on at least one of these paths?
$\textbf{... | (Anyone mind making a diagram for this)
Notice that the bug is basically moving from A to B (length 10) but going on a detour in the middle.
Specifically the detour would be of length 5 to some point and then going back by retracing its path (also length 5). Just to simplify things we can observe that the coordinates d... | 195 |
4,719 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_1 | 1 | What is $\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)$
$\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020$ | $20-2010+201+2010-201+20=20+20=\boxed{40}$ | 40 |
4,720 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_2 | 1 | A ferry boat shuttles tourists to an island every hour starting at 10 AM until its last trip, which starts at 3 PM. One day the boat captain notes that on the 10 AM trip there were 100 tourists on the ferry boat, and that on each successive trip, the number of tourists was 1 fewer than on the previous trip. How many to... | It is easy to see that the ferry boat takes $6$ trips total. The total number of people taken to the island is
\begin{align*}&100+99+98+97+96+95\\ &=6(100)-(1+2+3+4+5)\\ &=600 - 15\\ &=\boxed{585} | 585 |
4,721 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_3 | 1 | Rectangle $ABCD$ , pictured below, shares $50\%$ of its area with square $EFGH$ . Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$ . What is $\frac{AB}{AD}$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$ | From the problem statement, we know that \[\frac{[ABCD]}{2} = \frac{[EFGH]}{5} \Rightarrow [ABCD]=\frac{2[EFGH]}{5}\]
If we let $a = EF$ and $b = AD$ , we see \[[ABCD] = 2ab = \frac{2a^2}{5} \Rightarrow b = \frac{a}{5}\] . Hence, $\frac{AB}{AD} = \frac{2a}{b} = 2a(\frac{5}{a}) = \boxed{10}$ | 10 |
4,722 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_5 | 1 | Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next $n$ shots are bullseyes she will be guaranteed victory. What is the minimu... | Let $k$ be the number of points Chelsea currently has. In order to guarantee victory, we must consider the possibility that the opponent scores the maximum amount of points by getting only bullseyes.
\begin{align*}k+ 10n + 4(50-n) &> (k-50) + 50\cdot{10}\\ 6n &> 250\end{align*}
The lowest integer value that satisfies ... | 42 |
4,723 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_6 | 1 | $\text{palindrome}$ , such as $83438$ , is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qqu... | $x$ is at most $999$ , so $x+32$ is at most $1031$ . The minimum value of $x+32$ is $1000$ . However, the only palindrome between $1000$ and $1032$ is $1001$ , which means that $x+32$ must be $1001$
It follows that $x$ is $969$ , so the sum of the digits is $\boxed{24}$ | 24 |
4,724 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_8 | 1 | Triangle $ABC$ has $AB=2 \cdot AC$ . Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$ , respectively, such that $\angle BAE = \angle ACD$ . Let $F$ be the intersection of segments $AE$ and $CD$ , and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$
$\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ ... | Let $\angle BAE = \angle ACD = x$
\begin{align*}\angle BCD &= \angle AEC = 60^\circ\\ \angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\ \angle EAC &= 60^\circ - x\\ \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}
Since $\frac{AC}{A... | 90 |
4,725 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_9 | 1 | A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \q... | Imagine making the cuts one at a time. The first cut removes a box $2\times 2\times 3$ . The second cut removes two boxes, each of dimensions $2\times 2\times 0.5$ , and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$
Therefore the volume of ... | 7 |
4,726 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_9 | 2 | A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \q... | We can use Principle of Inclusion-Exclusion (PIE) to find the final volume of the cube.
There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has $2 \times 2 \times 3=12$ cubic inches. However, we can not just sum their volumes, as
the central $2\times 2\times 2$ cube is included i... | 7 |
4,727 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_9 | 3 | A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \q... | We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.
Each edge can be seen as a $2\times 0.5\times 0.5$ box, and each corner can be seen as a $0.5\times 0.5\times 0.5$ box.
$12\cdot{\frac{1}{2}} + 8\cdot{\frac{1}{8}} = 6+1 = \bo... | 7 |
4,728 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_10 | 1 | The first four terms of an arithmetic sequence are $p$ $9$ $3p-q$ , and $3p+q$ . What is the $2010^\text{th}$ term of this sequence?
$\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049$ | $3p-q$ and $3p+q$ are consecutive terms, so the common difference is $(3p+q)-(3p-q) = 2q$
\begin{align*}p+2q &= 9\\ 9+2q &= 3p-q\\ q&=2\\ p&=5\end{align*}
The common difference is $4$ . The first term is $5$ and the $2010^\text{th}$ term is
\[5+4(2009) = \boxed{8041}\] | 41 |
4,729 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_10 | 2 | The first four terms of an arithmetic sequence are $p$ $9$ $3p-q$ , and $3p+q$ . What is the $2010^\text{th}$ term of this sequence?
$\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049$ | Since all the answer choices are around $2010 \cdot 4 = 8040$ , the common difference must be $4$ . The first term is therefore $9 - 4 = 5$ , so the $2010^\text{th}$ term is $5 + 4 \cdot 2009 = \boxed{8041}$ | 41 |
4,730 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_12 | 1 | In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "Le... | Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
As Mike is a frog, his statement is false, hence there is at most one toad.
As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the... | 3 |
4,731 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_12 | 2 | In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "Le... | Notice that one of Chris and LeRoy must be a frog: if Chris is a frog, then he lies about LeRoy being a frog. Hence LeRoy is a toad. Alternatively, if Chris is a toad, then he tells the truth about LeRoy being a frog.
Assume Brian is a toad. Then Mike is a frog, and he lies about at least two being toads. This means th... | 3 |
4,732 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13 | 1 | For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$ | The image below shows the two curves for $k=4$ . The blue curve is $x^2+y^2=k^2$ , which is clearly a circle with radius $k$ , and the red curve is a part of the curve $xy=k$
[asy] import graph; size(200); real f(real x) {return 4/x;}; real g1(real x) {return sqrt(4*4-x*x);}; real g2(real x) {return -sqrt(4*4-x*x);}; ... | 2 |
4,733 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13 | 2 | For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$ | From the graph shown above, we see that there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the graph of the circle. This should be enough to conclude that the hyperbola does not intersect the circle.
Therefore, for each val... | 2 |
4,734 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13 | 3 | For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$ | Since $xy=k$ , multiply the equation by 2 on both sides to get $2xy=2k$ . Now we can add the two equations to get $(x+y)^2=k^2+2k$ , for which the only value of $k$ that does not satisfy the equation is $-1$ , as that makes the RHS negative. Similarly, if we subtract the two equations, we obtain $(x-y)^2=k^2-2k$ , for ... | 2 |
4,735 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13 | 4 | For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$ | Multiply $k=xy$ by and substitute it into $k^2=x^2+y^2$ . Then, $k=\frac{x^2+y^2}{xy}$ . Recognize it? It's also $k=\frac{x}{y}+\frac{y}{x}$ . The minimum of this function (more accurately the minimum absolute value of the function) is k=2, -2 (when x=y or x=-y). As long as k>2 or k<-2, the function is valid. As such, ... | 2 |
4,736 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13 | 5 | For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$ | Assume that $k\ge 0$ since if $k$ works then $-k$ also works. Let $x = ka$ and $y = kb$ . Then the given equations become $ab = \frac{1}{k}$ and $a^2 + b^2 = 1$ , which we don't want to intersect. The points that are the closest on this graph are $(1/\sqrt{2}, 1/\sqrt{2})$ and $(1/\sqrt{k})(1/\sqrt{k})$ . We don't want... | 2 |
4,737 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13 | 6 | For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$ | To know when the system does not intersect, we first need to find when they do intersect. To do so, we substitute $xy=k$ into $x^{2}+y^{2}=k^{2}$ giving:
$x^2+y^2=(xy)^2$
By Simon's Favorite Factoring Trick, we get $1=(1-x)(1+x)(1-y)(1+y)$
Notice that other than for $x$ and/or $y$ equal to $-1$ and/or $1$ because for a... | 2 |
4,738 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13 | 7 | For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$ | The first equation is describing a circle with radius $k$ centered at $(0,0)$ , so we can rewrite it in terms of a new variable $\theta$
\[(k\cos(\theta))^2+(k\sin(\theta))^2=k^2.\]
A quick check shows that this identity holds since $\sin^2+\cos^2=1$ . Now, with an alternate expression for $x$ and $y$ , we can plug thi... | 2 |
4,739 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_14 | 1 | Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3$ , and $DC=8$ . What is the smallest possible value of the perimeter?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$ | By the Angle Bisector Theorem , we know that $\frac{AB}{BC} = \frac{3}{8}$ . If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$ , respectively), then $AB + BC = AD + DC = AC$ , contradicting the Triangle Inequality . If we use the next lowest values ( $AB = 6$ and $BC = 16$ ), ... | 33 |
4,740 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_17 | 1 | Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$ . The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qqua... | It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$ . Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.
If we extend $BC$ $DE$ and $FA$ so that $F... | 6 |
4,741 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_18 | 1 | A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$ -coordinate or the $y$ -coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$ $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qqua... | Each path must go through either the second or the fourth quadrant.
Each path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$ $(-3,3)$ , and $(-2,2)$
There is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the t... | 698 |
4,742 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_18 | 2 | A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$ -coordinate or the $y$ -coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$ $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qqua... | As stated in the solution, there are $6$ points along the line $y=-x$ that constitute a sort of "boundary". Once the ant reaches one of these $6$ points, it is exactly halfway to $(4, 4)$ . Also notice that the ant will only cross one of the $6$ points during any one of its paths. Therefore we can divide the problem in... | 698 |
4,743 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_20 | 1 | Arithmetic sequences $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1<a_2 \le b_2$ and $a_n b_n = 2010$ for some $n$ . What is the largest possible value of $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 288 \qquad \textbf{(E)}\ 2009$ | Since $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1$ , we can write the terms of each sequence as
\begin{align*}&\left(a_n\right) \Rightarrow \{1, x+1, 2x+1, 3x+1, ...\}\\ &\left(b_n\right) \Rightarrow \{1, y+1, 2y+1, 3y+1, ...\}\end{align*}
where $x$ and $y$ $x\leq y$ ) are the common di... | 8 |
4,744 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_20 | 2 | Arithmetic sequences $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1<a_2 \le b_2$ and $a_n b_n = 2010$ for some $n$ . What is the largest possible value of $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 288 \qquad \textbf{(E)}\ 2009$ | Since
$a_n*b_n = 2010,$
and
$a_n \le b_n$ ,
blue+yellow=green
it follows that
$a_n \le \sqrt{2010} \Rightarrow a_n \le 44$
But $a_n$ and $b_n$ are also integers, so $a_n$ must be a factor of $2010$ smaller than $44$ . Notice that $2010 = 2*3*5*67$ . Therefore $a_n = 2, 3, 5, 6, 112, 15,$ or $30$ and $b_n = 1005, 670, 4... | 8 |
4,745 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_21 | 1 | The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$ , where the graph and the line intersect. What is the largest of these values?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$ | The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$
Since there are $3$ zeros and the function is never negative, all $3$ zeros must be double roots because the function's degree is $6$
Suppose we let $p$ $q$ , and $r$ be the roots ... | 4 |
4,746 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_21 | 2 | The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$ , where the graph and the line intersect. What is the largest of these values?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$ | The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$ .
We also know that this graph has 3 places tangent to the x-axis, which means that each root has to have a multiplicity of 2.
Let the function be $(x-p)^2(x-q)^2(x-r)^2$
Applying ... | 4 |
4,747 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_21 | 3 | The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$ , where the graph and the line intersect. What is the largest of these values?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$ | First, $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0$ has exactly $3$ roots. Therefore, $y = (kx^3+lx^2+mx+n)^2 = 0$
So, $k^2x^6+2klx^5+(2km+l^2)x^4+2(kn+lm)x^3+ax^2-bx-c = 0$
By matching the coefficients of the first $4$ terms, we have $k^2 = 1, 2kl = -10, 2km+l^2 = 29, 2kn+2lm = -4$
Solving the equations above, we have $2$ s... | 4 |
4,748 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_22 | 1 | What is the minimum value of $f(x)=\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$ | If we graph each term separately, we will notice that all of the zeros occur at $\frac{1}{m}$ , where $m$ is any integer from $1$ to $119$ , inclusive: $|mx-1|=0\implies mx=1\implies x=\frac{1}{m}$
The minimum value of $f(x)$ occurs where the absolute value of the sum of the slopes is at a minimum $\ge 0$ , since it is... | 49 |
4,749 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_22 | 2 | What is the minimum value of $f(x)=\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$ | By the triangle inequality, $|x-1|+|2x-1|+|3x-1|+\cdots + |119x-1| \geq |(x-1)+(2x-1)+\cdots+(119x)-1|.$ However, we may change signs of some of these terms to cancel out the $x$ 's.
Since the minimum exists, we want all the $x$ s to cancel out. Thus, we want to find some $n$ such that \[1+2+3+...+n=(n+1)+(n+2)+(n+3)+.... | 49 |
4,750 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_24 | 1 | Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ ... | The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.
We note that since all of the $\sin$ factors are inside a logarithm, the function is undefined where the inside of the logarithm is less than or equal ... | 12 |
4,751 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_24 | 2 | Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ ... | Note that the expression $\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)$ must be greater than zero, since logarithm functions are undefined for $0$ and negative numbers. Let $x_1, x_2, x_3, ..., x_8$ temporarily be the dependent variables of the functions $y_1 = \sin(\pi x_1), y_2 = \sin(2\p... | 12 |
4,752 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_24 | 3 | Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ ... | You should be able to somewhat visualize what the $\sin$ function looks like (if you can't then you should look it up and try to memorize it). To summarize, the graph of $y= \sin{x}$ is positive from the interval $(0, \pi)$ and negative from the interval $(\pi, 2\pi)$ (notice how the intervals use parentheses instead o... | 12 |
4,753 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_25 | 1 | Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad ... | It should first be noted that given any quadrilateral of fixed side lengths, there is exactly one way to manipulate the angles so that the quadrilateral becomes cyclic.
Proof. Given a quadrilateral $ABCD$ where all sides are fixed (in a certain order), we can construct the diagonal $\overline{BD}$ . When $BD$ is the mi... | 568 |
4,754 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_25 | 2 | Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad ... | As with solution $1$ we would like to note that given any quadrilateral we can change its angles to make a cyclic one.
Let $a \ge b \ge c\ge d$ be the sides of the quadrilateral.
There are $\binom{31}{3}$ ways to partition $32$ . However, some of these will not be quadrilaterals since they would have one side bigger t... | 568 |
4,755 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_25 | 3 | Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad ... | As with solution $1$ we find that there are $2255$ ways to form a quadrilateral if we don't account for rotations. We now apply Burnside's_Lemma . There are four types of actions in the group acting on the set of quadrilaterals. We will consider each individually:
Identity: maps a quadrilateral with sides $a,b,c,d$ in ... | 568 |
4,756 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_1 | 1 | Makarla attended two meetings during her $9$ -hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35$ | The total number of minutes in her $9$ -hour work day is $9 \times 60 = 540.$ The total amount of time spend in meetings in minutes is $45 + 45 \times 2 = 135.$ The answer is then $\frac{135}{540}$ $= \boxed{25}$ | 25 |
4,757 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_4 | 1 | A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | $31 \equiv 3 \pmod {7}$ so the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays. Therefore, Monday, Thursday, and Friday are valid so the answer is $\boxed{3}$ | 3 |
4,758 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_6 | 1 | At the beginning of the school year, $50\%$ of all students in Mr. Well's class answered "Yes" to the question "Do you love math", and $50\%$ answered "No." At the end of the school year, $70\%$ answered "Yes" and $30\%$ answered "No." Altogether, $x\%$ of the students gave a different answer at the beginning and end o... | Clearly, the minimum possible value would be $70 - 50 = 20\%$ . The maximum possible value would be $30 + 50 = 80\%$ . The difference is $80 - 20 = \boxed{60}$ | 60 |
4,759 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_8 | 1 | Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. Ho... | There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$ , because there wouldn't be a $64$ th place if x was. $x$ cannot ... | 23 |
4,760 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_8 | 2 | Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. Ho... | Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$
Since $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$ . This implies that $2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}$
Als... | 23 |
4,761 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_9 | 1 | Let $n$ be the smallest positive integer such that $n$ is divisible by $20$ $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$ | We know that $n^2 = k^3$ and $n^3 = m^2$ . Cubing and squaring the equalities respectively gives $n^6 = k^9 = m^4$ . Let $a = n^6$ . Now we know $a$ must be a perfect $36$ -th power because $lcm(9,4) = 36$ , which means that $n$ must be a perfect $6$ -th power. The smallest number whose sixth power is a multiple of $20... | 7 |
4,762 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_12 | 1 | For what value of $x$ does
\[\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?\]
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024$ | Using the fact that $\log_{x^n}{y^n} = \log_{x}{y}$ , we see that the equation becomes $\log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} = 40$ . Thus, $5\log_{2}{x} = 40$ and $\log_{2}{x} = 8$ , so $x = 2^8 = 256$ , or $\boxed{256}$ | 256 |
4,763 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_14 | 1 | Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$ | We want to try make $a+b$ $b+c$ $c+d$ , and $d+e$ as close as possible so that $M$ , the maximum of these, is smallest.
Notice that $2010=670+670+670$ . In order to express $2010$ as a sum of $5$ numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as clos... | 671 |
4,764 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_14 | 2 | Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$ | Since $a + b \le M$ $d + e \le M$ , and $c < b + c \le M$ , we have that $2010 = a + b + c + d + e < 3M$ . Hence, $M > 670$ , or $M \ge 671$
For the values $(a,b,c,d,e) = (669,1,670,1,669)$ $M = 671$ , so the smallest possible value of $M$ is $\boxed{671}$ . The answer is (B). | 671 |
4,765 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_14 | 3 | Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$ | Notice that only the sums of adjacent numbers matter. (For example, a & c could be extremely high, as long as b is relatively low.)
Therefore creating "mountains" and "valleys" is the best way to lower the sum of adjacent numbers. We can do
1. (high, low, high, low, high)
or
2. (low, high, low, high, low)
In the extrem... | 671 |
4,766 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_15 | 1 | For how many ordered triples $(x,y,z)$ of nonnegative integers less than $20$ are there exactly two distinct elements in the set $\{i^x, (1+i)^y, z\}$ , where $i=\sqrt{-1}$
$\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235$ | We have either $i^{x}=(1+i)^{y}\neq z$ $i^{x}=z\neq(1+i)^{y}$ , or $(1+i)^{y}=z\neq i^x$
$i^{x}=(1+i)^{y}$ only occurs when it is $1$ $(1+i)^{y}=1$ has only one solution, namely, $y=0$ $i^{x}=1$ has five solutions between 0 and 19, $x=0, x=4, x=8, x=12$ , and $x=16$ $z\neq 1$ has nineteen integer solutions between zero... | 225 |
4,767 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_17 | 1 | The entries in a $3 \times 3$ array include all the digits from $1$ through $9$ , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$ | Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right.
\[\begin{tabular}{|c|c|c|} \hline 1&2&\\... | 42 |
4,768 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_17 | 2 | The entries in a $3 \times 3$ array include all the digits from $1$ through $9$ , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$ | This solution is trivial by the hook length theorem. The hooks look like this:
$\begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\ \hline 4 & 3 & 2\\ \hline 3 & 2 & 1\\ \hline \end{tabular}$
So, the answer is $\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}$ $\boxed{42}$ | 42 |
4,769 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_19 | 1 | A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arit... | Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know $r > 1$ because the sequence is said to be increasing. We also know that each of $a, ar, ar^2, ar^3$ is an integer. We start by showing that $r$ must also be an integer.
Suppose not, and say $r = m/n$ where $m>n>1$ , and $\gcd(m,n)=1$ . Then $n, ... | 34 |
4,770 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_19 | 2 | A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arit... | Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let $a,a+d,a+2d,a+3d$ be the quarterly scores for the Wildcats. The sum of the Raiders scores is $a(1+r+r^{2}+r^{3})$ and the sum of the Wildcats scores i... | 34 |
4,771 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_20 | 1 | A geometric sequence $(a_n)$ has $a_1=\sin x$ $a_2=\cos x$ , and $a_3= \tan x$ for some real number $x$ . For what value of $n$ does $a_n=1+\cos x$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$ | By the defintion of a geometric sequence, we have $\cos^2x=\sin x \tan x$ . Since $\tan x=\frac{\sin x}{\cos x}$ , we can rewrite this as $\cos^3x=\sin^2x$
The common ratio of the sequence is $\frac{\cos x}{\sin x}$ , so we can write
\[a_1= \sin x\] \[a_2= \cos x\] \[a_3= \frac{\cos^2x}{\sin x}\] \[a_4=\frac{\cos^3x}{\... | 8 |
4,772 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_20 | 2 | A geometric sequence $(a_n)$ has $a_1=\sin x$ $a_2=\cos x$ , and $a_3= \tan x$ for some real number $x$ . For what value of $n$ does $a_n=1+\cos x$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$ | Notice that the common ratio is $r=\frac{\cos(x)}{\sin(x)}$ ; multiplying it to $\tan(x)=\frac{\sin(x)}{\cos(x)}$ gives $a_4=1$ . Then, working backwards we have $a_3=\frac{1}{r}$ $a_2=\frac{1}{r^2}$ and $a_1=\frac{1}{r^3}$ . Now notice that since $a_1=\sin(x)$ and $a_2=\cos(x)$ , we need $a_1^2+a_2^2=1$ , so $\frac{1}... | 8 |
4,773 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_21 | 1 | Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that
What is the smallest possible value of $a$
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$ | We observe that because $P(1) = P(3) = P(5) = P(7) = a$ , if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$ $R(x)$ has roots when $P(x) = a$ ; namely, when $x=1,3,5,7$
Thus since $R(x)$ has roots when $x=1,3,5,7$ , we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial ... | 315 |
4,774 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_21 | 2 | Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that
What is the smallest possible value of $a$
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$ | The evenly-spaced data suggests using discrete derivatives to tackle this problem. First, note that any polynomial of degree $n$
can also be written as
Moreover, the coefficients $a_i$ are integers for $i=1, 2, \ldots n$ iff the coefficients $b_i$ are integers for $i=1, 2, \ldots n$ . This latter form is convenient f... | 315 |
4,775 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_23 | 1 | Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$ , and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$ . What is the sum of the minimum values of $P(x)$ and $Q(x)$
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf... | $P(x) = (x - a)^2 - b, Q(x) = (x - c)^2 - d$ . Notice that $P(x)$ has roots $a\pm \sqrt {b}$ , so that the roots of $P(Q(x))$ are the roots of $Q(x) = a + \sqrt {b}, a - \sqrt {b}$ . For each individual equation, the sum of the roots will be $2c$ (symmetry or Vieta's). Thus, we have $4c = - 23 - 21 - 17 - 15$ , or $c =... | 100 |
4,776 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_23 | 3 | Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$ , and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$ . What is the sum of the minimum values of $P(x)$ and $Q(x)$
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf... | Let $P(x) = x^2 - (a+b)x + ab$ and $Q(x) = x^2 - (c+d)x + cd$ . Notice that the roots of $P(x)$ are $a,b$ and the roots of $Q(x)$ are $c,d.$ Then we get:
\begin{align*} P(Q(x)) &= a, b \\ x^2 - (c+d)x + cd &= a, b \end{align*} The two possible equations are then $x^2 - (c+d)x + cd-a=0$ and $x^2 - (c+d)x + cd-b=0$ . The... | 100 |
4,777 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_23 | 4 | Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$ , and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$ . What is the sum of the minimum values of $P(x)$ and $Q(x)$
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf... | Let $P(x) = (x+a)(x+b)$ $Q(x) = (x+c)(x+d)$
$P(Q(x)) = (x^2 + cx + dx + cd + a)(x^2 + cx + dx + cd + b)$
$Q(P(x)) = (x^2 + ax + bx + ab + c)(x^2 + ax + bx + ab + d)$
Notice how the coefficient for $x$ has to be the same for the two quadratics that are multiplied to create $P(Q(x))$ , and $Q(P(x))$
$P(Q(x)) = (x+ 23)(x+... | 100 |
4,778 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_25 | 1 | For every integer $n\ge2$ , let $\text{pow}(n)$ be the largest power of the largest prime that divides $n$ . For example $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$ . What is the largest integer $m$ such that $2010^m$ divides
$\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \q... | Because 67 is the largest prime factor of 2010, it means that in the prime factorization of $\prod_{n=2}^{5300}\text{pow}(n)$ , there'll be $p_1 ^{e_1} \cdot p_2 ^{e_2} \cdot .... 67^x ...$ where $x$ is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times $67$ is inco... | 77 |
4,779 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_25 | 2 | For every integer $n\ge2$ , let $\text{pow}(n)$ be the largest power of the largest prime that divides $n$ . For example $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$ . What is the largest integer $m$ such that $2010^m$ divides
$\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \q... | After finding the prime factorization of $2010=2\cdot3\cdot5\cdot67$ , divide $5300$ by $67$ and add $5300$ divided by $67^2$ in order to find the total number of multiples of $67$ between $2$ and $5300$ $\lfloor\frac{5300}{67}\rfloor+\lfloor\frac{5300}{67^2}\rfloor=80$ Since $71$ $73$ , and $79$ are prime numbers grea... | 77 |
4,780 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_4 | 1 | Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 45 \qquad \textbf{(E)}\ 55$ | As all five options are divisible by $5$ , we may not use any pennies. (This is because a penny is the only coin that is not divisible by $5$ , and if we used between $1$ and $4$ pennies, the sum would not be divisible by $5$ .)
Hence the smallest coin we can use is a nickel, and thus the smallest amount we can get is ... | 15 |
4,781 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_5 | 1 | One dimension of a cube is increased by $1$ , another is decreased by $1$ , and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 125 \qquad \textb... | Let the original cube have edge length $a$ . Then its volume is $a^3$ .
The new box has dimensions $a-1$ $a$ , and $a+1$ , hence its volume is $(a-1)a(a+1) = a^3-a$ .
The difference between the two volumes is $a$ . As we are given that the difference is $5$ , we have $a=5$ , and the volume of the original cube was $5^... | 125 |
4,782 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_6 | 1 | Suppose that $P = 2^m$ and $Q = 3^n$ . Which of the following is equal to $12^{mn}$ for every pair of integers $(m,n)$
$\textbf{(A)}\ P^2Q \qquad \textbf{(B)}\ P^nQ^m \qquad \textbf{(C)}\ P^nQ^{2m} \qquad \textbf{(D)}\ P^{2m}Q^n \qquad \textbf{(E)}\ P^{2n}Q^m$ | We have $12^{mn} = (2\cdot 2\cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = \boxed{2}$ | 2 |
4,783 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_7 | 1 | The first three terms of an arithmetic sequence are $2x - 3$ $5x - 11$ , and $3x + 1$ respectively. The $n$ th term of the sequence is $2009$ . What is $n$
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 502 \qquad \textbf{(C)}\ 1004 \qquad \textbf{(D)}\ 1506 \qquad \textbf{(E)}\ 8037$ | As this is an arithmetic sequence, the difference must be constant: $(5x-11) - (2x-3) = (3x+1) - (5x-11)$ . This solves to $x=4$ . The first three terms then are $5$ $9$ , and $13$ . In general, the $n$ th term is $1+4n$ . Solving $1+4n=2009$ , we get $n=\boxed{502}$ | 502 |
4,784 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_8 | 1 | Four congruent rectangles are placed as shown. The area of the outer square is $4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ 2 + \sqrt2 \qquad \textbf{(D)}\... | Let the side length of the smaller square be $1$ , and let the smaller side of the rectangles be $y$ . Since the larger square's area is four times larger than the smaller square's, the larger square's side length is $2$ $2$ is equivalent to $2y+1$ , giving $y=1/2$ . Then, the longer side of the rectangles is $3/2$ $\f... | 3 |
4,785 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_9 | 1 | Suppose that $f(x+3)=3x^2 + 7x + 4$ and $f(x)=ax^2 + bx + c$ . What is $a+b+c$
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$ | As $f(x)=ax^2 + bx + c$ , we have $f(1)=a\cdot 1^2 + b\cdot 1 + c = a+b+c$
To compute $f(1)$ , set $x=-2$ in the first formula. We get $f(1) = f(-2+3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = \boxed{2}$ | 2 |
4,786 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_9 | 2 | Suppose that $f(x+3)=3x^2 + 7x + 4$ and $f(x)=ax^2 + bx + c$ . What is $a+b+c$
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$ | Combining the two formulas, we know that $f(x+3) = a(x+3)^2 + b(x+3) + c$
We can rearrange the right hand side to $ax^2 + (6a+b)x + (9a+3b+c)$
Comparing coefficients we have $a=3$ $6a+b=7$ , and $9a+3b+c = 4$ . From the second equation we get $b=-11$ , and then from the third we get $c=10$ . Hence $a+b+c = 3-11+10 = \b... | 2 |
4,787 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_10 | 1 | In quadrilateral $ABCD$ $AB = 5$ $BC = 17$ $CD = 5$ $DA = 9$ , and $BD$ is an integer. What is $BD$
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$ | By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$ , and also $BD + CD > BC$ , hence $BD > BC - CD = 17 - 5 = 12$
We get that $12 < BD < 14$ , and as we know that $BD$ is an integer, we must have $BD=\boxed{13}$ | 13 |
4,788 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_11 | 1 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13... | Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$ th triangular number (i.e. the diamonds within the triangles or between the diagonals) and $4(n-1)+1$ (the diamonds on sides of the triang... | 761 |
4,789 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_11 | 2 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13... | Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle; marker mred=marker(scale(5)*d,red,Fill); marker mblue=marker(scale(5)*d,blue,Fi... | 761 |
4,790 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_11 | 3 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13... | When constructing $F_n$ from $F_{n-1}$ , we add $4(n-1)$ new diamonds. Let $d_n$ be the number of diamonds in $F_n$ . We now know that $d_1=1$ and $\forall n>1:~ d_n=d_{n-1} + 4(n-1)$
Hence we get: \begin{align*} d_{20} & = d_{19} + 4\cdot 19 \\ & = d_{18} + 4\cdot 18 + 4\cdot 19 \\ & = \cdots \\ & = 1 + 4(1+2+\cdots+1... | 761 |
4,791 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_11 | 4 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13... | The sequence $\{ d_n\}$ goes $1, 5, 13, 25, 41,\dots$ . The first finite differences go $4, 8, 12, 16, \dots$ . The second finite differences go $4, 4, 4, \dots$ , so we see that the second finite difference is constant. Thus, $d_n$ can be represented as a quadratic, $d_n = an^2 + bn + c$ . However, we already know $d_... | 761 |
4,792 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_12 | 2 | How many positive integers less than $1000$ are $6$ times the sum of their digits?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 12$ | The sum of the digits is at most $9+9+9=27$ . Therefore the number is at most $6\cdot 27 = 162$ . Since the number is $6$ times the sum of its digits, it must be divisible by $6$ , therefore also by $3$ , therefore the sum of its digits must be divisible by $3$ . With this in mind we can conclude that the number must b... | 1 |
4,793 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_15 | 1 | For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$
Note: here $i = \sqrt { - 1}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$ | We know that $i^x$ cycles every $4$ powers so we group the sum in $4$ s. \[i+2i^2+3i^3+4i^4=2-2i\] \[5i^5+6i^6+7i^7+8i^8=2-2i\]
We can postulate that every group of $4$ is equal to $2-2i$ .
For 24 groups we thus, get $48-48i$ as our sum.
We know the solution must lie near
The next term is the $24*4+1=97$ th term. This... | 97 |
4,794 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_15 | 2 | For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$
Note: here $i = \sqrt { - 1}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$ | Obviously, even powers of $i$ are real and odd powers of $i$ are imaginary.
Hence the real part of the sum is $2i^2 + 4i^4 + 6i^6 + \ldots$ , and
the imaginary part is $i + 3i^3 + 5i^5 + \cdots$
Let's take a look at the real part first. We have $i^2=-1$ , hence the real part simplifies to $-2+4-6+8-10+\cdots$ .
If th... | 97 |
4,795 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_15 | 3 | For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$
Note: here $i = \sqrt { - 1}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$ | Some may know the equation:
\[\sum_{k=1}^{n}kr^{k-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}\]
(For those curious, this comes from differentiating the equation for finite geometric sums)
Using this equation, we have
\[48+49i=i\frac{1-(n+1)i^n+ni^{n+1}}{(1-i)^2}\] \[=\frac{1-(n+1)i^n+ni^{n+1}}{-2}\] \[=-\frac{1}{2}+\frac{(n... | 97 |
4,796 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_16 | 1 | A circle with center $C$ is tangent to the positive $x$ and $y$ -axes and externally tangent to the circle centered at $(3,0)$ with radius $1$ . What is the sum of all possible radii of the circle with center $C$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ ... | Let $r$ be the radius of our circle. For it to be tangent to the positive $x$ and $y$ axes, we must have $C=(r,r)$ . For the circle to be externally tangent to the circle centered at $(3,0)$ with radius $1$ , the distance between $C$ and $(3,0)$ must be exactly $r+1$
By the Pythagorean theorem the distance between $(r,... | 8 |
4,797 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_17 | 1 | Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1$ , and the sum of the second series is $r_2$ . What is $r_1 + r_2$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac... | Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$
Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ .
This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$
As we are given that $r_1$ and $r_2$ are distinct, these ... | 1 |
4,798 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_17 | 2 | Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1$ , and the sum of the second series is $r_2$ . What is $r_1 + r_2$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac... | Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$
Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ .
This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$
Which can be further rewritten as $r_1-r_1^2 = r_2-r_2^2$... | 1 |
4,799 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_18 | 1 | For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 1... | The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$
For $k\in\{1,2,3\}$ we have $I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)$ . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\leq 5$
For $k>4$ we have $I_k=2^6 \left( 5^{k+2}\... | 7 |
4,800 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_18 | 2 | For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 1... | Notice that $2$ is a prime factor of an integer $n$ if and only if $n$ is even. Therefore, given any sufficiently high positive integral value of $k$ , dividing $I_k$ by $2^6$ yields a terminal digit of zero, and dividing by 2 again leaves us with $2^7 \cdot a = I_k$ where $a$ is an odd integer.
Observe then that $\box... | 7 |
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