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 |
|---|---|---|---|---|---|
6,301 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_3 | 2 | Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer. | Let $N$ be the required number, and $N'$ be $N$ with the first digit deleted. Now, we know that $N<1000$ (because this is an AIME problem). Thus, $N$ has $1,$ $2$ or $3$ digits. Checking the other cases, we see that it must have $3$ digits.
Let $N=\overline{abc}$ , so $N=100a+10b+c$ . Thus, $N'=\overline{bc}=10b+c$ . B... | 725 |
6,302 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_4 | 1 | Let $N$ be the number of consecutive $0$ 's at the right end of the decimal representation of the product $1!2!3!4!\cdots99!100!.$ Find the remainder when $N$ is divided by $1000$ | A number in decimal notation ends in a zero for each power of ten which divides it. Thus, we need to count both the number of 5s and the number of 2s dividing into our given expression. Since there are clearly more 2s than 5s, it is sufficient to count the number of 5s.
One way to do this is as follows: $96$ of the n... | 124 |
6,303 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_5 | 1 | The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers . Find $abc$ | We begin by equating the two expressions:
\[a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}\]
Squaring both sides yields:
\[2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006\]
Since $a$ $b$ , and $c$ are integers, we can match c... | 936 |
6,304 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_5 | 2 | The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers . Find $abc$ | We realize that the quantity under the largest radical is a perfect square and attempt to rewrite the radicand as a square. Start by setting $x=\sqrt{2}$ $y=\sqrt{3}$ , and $z=\sqrt{5}$ . Since
\[(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)\]
we attempt to rewrite the radicand in this form:
\[2006+2(52xy+234xz+... | 936 |
6,305 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_6 | 1 | Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits . Find the sum of the elements of $\mathcal{S}.$ | Numbers of the form $0.\overline{abc}$ can be written as $\frac{abc}{999}$ . There are $10\times9\times8=720$ such numbers. Each digit will appear in each place value $\frac{720}{10}=72$ times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is $\frac{45\times72\times111}{999}= \boxed{360}$ | 360 |
6,306 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_6 | 2 | Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits . Find the sum of the elements of $\mathcal{S}.$ | Alternatively, for every number, $0.\overline{abc}$ , there will be exactly one other number, such that when they are added together, the sum is $0.\overline{999}$ , or, more precisely, 1. As an example, $.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1$
Thus, the solution can be determined by dividing... | 360 |
6,307 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_7 | 1 | An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,... | Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof
Let the set of parallel lines be perpendicular to the x-axis , such that they cross it at $0, 1, 2 \ldots$ . The base of region $\mathcal{A}$ is on the line $x = 1$ . The bigger base of region $\mathcal{D}$ is on the line $x = 7$ . ... | 408 |
6,308 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_7 | 3 | An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,... | Let the distances from the apex to the parallel lines be $x$ and $y$ and the distance between the intersections be $a,b.$ We know the area ratio means $\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\frac{5}{11}$ which simplifying yields $ab = 3ay+3bx.$ The ratio we seek is $\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}... | 408 |
6,309 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_11 | 1 | A collection of 8 cubes consists of one cube with edge length $k$ for each integer $k, 1 \le k \le 8.$ A tower is to be built using all 8 cubes according to the rules:
Let $T$ be the number of different towers than can be constructed. What is the remainder when $T$ is divided by 1000? | We proceed recursively . Suppose we can build $T_m$ towers using blocks of size $1, 2, \ldots, m$ . How many towers can we build using blocks of size $1, 2, \ldots, m, m + 1$ ? If we remove the block of size $m + 1$ from such a tower (keeping all other blocks in order), we get a valid tower using blocks $1, 2, \ldot... | 458 |
6,310 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_12 | 1 | Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$ , where $x$ is measured in degrees and $100< x< 200.$ | Observe that $2\cos 4x\cos x = \cos 5x + \cos 3x$ by the sum-to-product formulas. Defining $a = \cos 3x$ and $b = \cos 5x$ , we have $a^3 + b^3 = (a+b)^3 \rightarrow ab(a+b) = 0$ . But $a+b = 2\cos 4x\cos x$ , so we require $\cos x = 0$ $\cos 3x = 0$ $\cos 4x = 0$ , or $\cos 5x = 0$
Hence we see by careful analysis of ... | 906 |
6,311 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13 | 1 | For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square | Given $g : x \mapsto \max_{j : 2^j | x} 2^j$ , consider $S_n = g(2) + \cdots + g(2^n)$ . Define $S = \{2, 4, \ldots, 2^n\}$ . There are $2^0$ elements of $S$ that are divisible by $2^n$ $2^1 - 2^0 = 2^0$ elements of $S$ that are divisible by $2^{n-1}$ but not by $2^n, \ldots,$ and $2^{n-1}-2^{n-2} = 2^{n-2}$ elements o... | 899 |
6,312 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13 | 2 | For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square | First note that $g(k)=1$ if $k$ is odd and $2g(k/2)$ if $k$ is even.
so $S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).$ $2k-1$ must be odd so this reduces to $2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).$ Thus $S_n=2(2... | 899 |
6,313 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13 | 3 | For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square | At first, this problem looks kind of daunting, but we can easily solve this problem by finding patterns, recursion and algebraic manipulations.
We first simplify all the messy notation in the $S_n$ term. Note that the problem asks us to find the smallest value of $n<1000$ such that there exists an integer $k$ that sati... | 899 |
6,314 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13 | 4 | For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square | First, we set intervals. Say that a number $N$ falls strictly within $[2^k, 2^{k+1}]$
$N<2^{k+1}=2^k+2^k$
We can generalize this:
If a number is in the form $N=2^k+2^{R}O$ where $O$ is a positive odd number, $R<k$
$N<2^{k+1}=2^k+2^k\Longrightarrow O<2^{k-R}$ so there are $2^{k-R-1}$ numbers that satisfy this form.
For... | 899 |
6,315 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_14 | 1 | A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top o... | Diagram borrowed from Solution 1
Apply Pythagorean Theorem on $\bigtriangleup TOB$ yields \[BO=\sqrt{TB^2-TO^2}=3\] Since $\bigtriangleup ABC$ is equilateral, we have $\angle MOB=60^{\circ}$ and \[BC=2BM=2(OB\sin MOB)=3\sqrt{3}\] Apply Pythagorean Theorem on $\bigtriangleup TMB$ yields \[TM=\sqrt{TB^2-BM^2}=\sqrt{5^2-(... | 183 |
6,316 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_2 | 1 | The lengths of the sides of a triangle with positive area are $\log_{10} 12$ $\log_{10} 75$ , and $\log_{10} n$ , where $n$ is a positive integer. Find the number of possible values for $n$ | By the Triangle Inequality and applying the well-known logarithmic property $\log_{c} a + \log_{c} b = \log_{c} ab$ , we have that
Also,
Combining these two inequalities:
\[6.25 < n < 900\]
Thus $n$ is in the set $(6.25 , 900)$ ; the number of positive integer $n$ which satisfies this requirement is $\boxed{893}$ | 893 |
6,317 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_3 | 2 | Let $P$ be the product of the first $100$ positive odd integers . Find the largest integer $k$ such that $P$ is divisible by $3^k .$ | We are obviously searching for multiples of three set S of odd numbers 1-199. Starting with 3, every number $\equiv 2 \pmod{3}$ in set S will be divisible by 3. In other words, every number $\equiv 3 \pmod{6}$ . This is because the LCM must be divisible by 3, and 2, because the set is comprised of only odd numbers. Usi... | 49 |
6,318 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_5 | 1 | When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on op... | Without loss of generality , assume that face $F$ has a 6, so the opposite face has a 1. Let $A(n)$ be the probability of rolling a number $n$ on one die and let $B(n)$ be the probability of rolling a number $n$ on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probabil... | 29 |
6,319 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_5 | 2 | When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on op... | We have that the cube probabilities to land on its faces are $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}+x$ $\frac{1}{6}-x$ we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: \[4 \cdot \left(\frac{1}{6} \right)^2+... | 29 |
6,320 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_5 | 3 | When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on op... | Let $p(a,b)$ denote the probability of obtaining $a$ on the first die and $b$ on the second. Then the probability of obtaining a sum of 7 is \[p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).\] Let the probability of obtaining face $F$ be $(1/6)+x$ . Then the probability of obtaining the face opposite face $F$ is $(1/6)-x$ .... | 29 |
6,321 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6 | 1 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqr... | [asy] unitsize(32mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1); pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1); pair Ap = (0, (3 - sqrt(3))/6); pair Cp = ((3 - sqrt(3))/6, 0); pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6); pair[] dots = {A, B, C, D, ... | 12 |
6,322 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6 | 2 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqr... | Suppose $\overline{AB} = \overline{AD} = x.$ Note that $\angle EAF = 60$ since the triangle is equilateral, and by symmetry, $\angle BAE = \angle DAF = 15.$ Note that if $\overline{AD} = x$ and $\angle BAE = 15$ , then $\overline{AA'}=\frac{x}{\tan(15)}.$ Also note that \[AB = 1 = \overline{AA'} + \overline{A'B} = \fra... | 12 |
6,323 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6 | 3 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqr... | Why not solve in terms of the side $x$ only (single-variable beauty)? By similar triangles we obtain that $BE=\frac{x}{1-x}$ , therefore $CE=\frac{1-2x}{1-x}$ . Then $AE=\sqrt{2}*\frac{1-2x}{1-x}$ . Using Pythagorean Theorem on $\triangle{ABE}$ yields $\frac{x^2}{(1-x)^2} + 1 = 2 * \frac{(1-2x)^2}{(1-x)^2}$ . This mean... | 12 |
6,324 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_7 | 1 | Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit. | There are $\left\lfloor\frac{999}{10}\right\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that ... | 738 |
6,325 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_7 | 2 | Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit. | We proceed by casework on the number of digits of $a.$
Case 1: Both $a$ and $b$ have three digits
We now use constructive counting. For the hundreds digit of $a,$ we see that there are $8$ options - the numbers $1$ through $8.$ (If $a = 9,$ that means that $b$ will be a two digit number, and if $a = 0,$ $a$ will have t... | 738 |
6,326 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_7 | 3 | Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit. | For every $a \in [1, 999]$ $(a, 1000 - a)$ is a potential candidate for a solution, barring the cases where $a$ or $1000 -a$ has zero digits.
First, let's consider all viable $a$ that do not have a zero digit. As there are 9 non-zero digits, we have:
However, we have still overlooked the cases where $1000 - a$ contains... | 738 |
6,327 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_8 | 1 | There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible t... | If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection. Thus, to make two triangles distinct, they must differ either in their center triangle or i... | 336 |
6,328 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_8 | 2 | There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible t... | We apply Burnside's Lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by $3+3=6$ for our final count.
Case 1: 0 degree rotation. This is known as the identity rotation, and there are $6^... | 336 |
6,329 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_8 | 3 | There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible t... | There are $6$ choices for the center triangle. Note that given any $3$ colors, there is a unique way to assign them to the corner triangles. We have $6$ different colors to choose from, so the number of ways to color the corner triangles is the same as the number of ways to arrange $6 - 1 = 5$ dividers and $3$ identica... | 336 |
6,330 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10 | 1 | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulate... | The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$ . We let this probability be $p$ ; then the probability that $A$ and $B$ end with the same score in t... | 831 |
6,331 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10 | 2 | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulate... | You can break this into cases based on how many rounds $A$ wins out of the remaining $5$ games.
Summing these 6 cases, we get $\frac{638}{1024}$ , which simplifies to $\frac{319}{512}$ , so our answer is $319 + 512 = \boxed{831}$ | 831 |
6,332 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10 | 3 | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulate... | We can apply the concept of generating functions here.
The generating function for $B$ is $(1 + 0x^{1})$ for the first game where $x^{n}$ is winning n games. Since $B$ lost the first game, the coefficient for $x^{1}$ is 0. The generating function for the next 5 games is $(1 + x)^{5}$ . Thus, the total generating functi... | 831 |
6,333 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10 | 4 | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulate... | After the first game, there are $10$ games we care about-- those involving $A$ or $B$ . There are $3$ cases of these $10$ games: $A$ wins more than $B$ $B$ wins more than $A$ , or $A$ and $B$ win the same number of games. Also, there are $2^{10} = 1024$ total outcomes. By symmetry, the first and second cases are equall... | 831 |
6,334 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_11 | 1 | sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000. | Define the sum as $s$ . Since $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$ , the sum will be:
Thus $s = \frac{a_{28} + a_{30}}{2}$ , and $a_{28},\,a_{30}$ are both given; the last four digits of their sum is $3668$ , and half of that is $1834$ . Therefore, the answer is $\boxed{834}$ .− | 834 |
6,335 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12 | 1 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overli... | Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$ . Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$ . So $\Delta{GBC} \sim \Delta{EAF}$
\[[EAF] = \frac12 (AE)(EF)\sin \angle AEF = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.\]... | 865 |
6,336 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12 | 2 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overli... | Solution by e_power_pi_times_i/edited by srisainandan6
Let the center of the circle be $O$ and the origin. Then, $A (0,2)$ $B (-\sqrt{3}, -1)$ $C (\sqrt{3}, -1)$ $D$ and $E$ can be calculated easily knowing $AD$ and $AE$ $D (-\dfrac{13}{2}, \dfrac{-13\sqrt{3}+4}{2})$ $E (\dfrac{11}{2}, \dfrac{-11\sqrt{3}+4}{2})$ . As $... | 865 |
6,337 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12 | 3 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overli... | Lines $l_1$ and $l_2$ are constructed such that $AEFD$ is a parallelogram, hence $DF = 13$ . Since $BAC$ is equilateral with angle of $60^{\circ}$ , angle $D$ is $120^{\circ}$ . Use law of cosines to find $AF = \sqrt{433}$ . Then use law of sines to find angle $BAG$ and $GAC$ . Next we use Ptolemy's Theorem on $ABGC$ t... | 865 |
6,338 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12 | 4 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overli... | Note that $AB=2\sqrt3$ $DF=11$ , and $EF=13$ . If we take a homothety of the parallelogram with respect to $A$ , such that $F$ maps to $G$ , we see that $\frac{[ABG]}{[ACG]}=\frac{11}{13}$ . Since $\angle AGB=\angle AGC=60^{\circ}$ , from the sine area formula we have $\frac{BG}{CG}=\frac{11}{13}$ . Let $BG=11k$ and $C... | 865 |
6,339 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_13 | 1 | How many integers $N$ less than $1000$ can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$ | Let the first odd integer be $2n+1$ $n\geq 0$ . Then the final odd integer is $2n+1 + 2(j-1) = 2(n+j) - 1$ . The odd integers form an arithmetic sequence with sum $N = j\left(\frac{(2n+1) + (2(n+j)-1)}{2}\right) = j(2n+j)$ . Thus, $j$ is a factor of $N$
Since $n\geq 0$ , it follows that $2n+j \geq j$ and $j\leq \sqrt{N... | 15 |
6,340 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_1 | 1 | Six congruent circles form a ring with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius 30. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $\lfloor K \rfloor$ (the floor function ).
2005 AIM... | Define the radii of the six congruent circles as $r$ . If we draw all of the radii to the points of external tangency, we get a regular hexagon . If we connect the vertices of the hexagon to the center of the circle $C$ , we form several equilateral triangles . The length of each side of the triangle is $2r$ . Notice t... | 942 |
6,341 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_3 | 1 | How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50? | Suppose $n$ is such an integer . Because $n$ has $3$ proper divisors, it must have $4$ divisors,, so $n$ must be in the form $n=p\cdot q$ or $n=p^3$ for distinct prime numbers $p$ and $q$
In the first case, the three proper divisors of $n$ are $1$ $p$ and $q$ . Thus, we need to pick two prime numbers less than $50$ . ... | 109 |
6,342 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4 | 1 | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no ... | If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$ . If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$ . Thus, if the number of columns is $n$ , the number of students is $n(n... | 294 |
6,343 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4 | 2 | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no ... | Let there be $m$ members and $n$ members for the square and $c$ for the number of columns of the other formation. We have $n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}... | 294 |
6,344 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4 | 3 | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no ... | Think of the process of moving people from the last column to new rows. Since there are less columns than rows, for each column removed, there are people discarded to the "extra" pile to be placed at the end. To maximize the number of "extra" people to fill in the last few rows. We remove 3 columns and add 4 rows. For ... | 294 |
6,345 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4 | 4 | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no ... | Note: Only do this if you have a LOT of time (and you've memorized all your perfect squares up to 1000).
We can see that the number of members in the band must be of the form $n(n + 7)$ for some positive integer $n$ . When $n = 28$ , this product is $980$ , and since AIME answers are nonnegative integers less than $100... | 294 |
6,346 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_5 | 1 | Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangeme... | There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.
There are ${8\choose4} = 70$ ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.
Create a string of letters H and T to denote the orientation of the... | 630 |
6,347 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_5 | 2 | Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangeme... | First, we can break this problem up into two parts: the amount of ways to order the coins based on color, and which side is facing up for each coin (assuming they are indistinguishable). In the end, we multiply these values together.
First, there are obviously $\binom{8}{4}$ ways to order the coins based on color.
Next... | 630 |
6,348 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_7 | 1 | In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers , find $p+q.$ | [asy]draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label("$A$",(0,0),SW); label("$B$",(20.87,0),SE); label("$E$",(15.87,8.66),NE); label("$D$",(5,8.66),NW); label("$P$",(5,0),S); label("$Q$",(15.87,0),S); label("$C$",(16.87,7)... | 150 |
6,349 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_7 | 2 | In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers , find $p+q.$ | Draw the perpendiculars from $C$ and $D$ to $AB$ , labeling the intersection points as $E$ and $F$ . This forms 2 $30-60-90$ right triangles , so $AE = 5$ and $BF = 4$ . Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$ , we find another right triangle $\triangle DGC$ $DG = DE - CF... | 150 |
6,350 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_8 | 1 | The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots . Given that their sum is $m/n$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$ | Let $y = 2^{111x}$ . Then our equation reads $\frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$ . Thus, if this equation has roots $r_1, r_2$ and $r_3$ , by Vieta's formulas we have $r_1\cdot r_2\cdot r_3 = 4$ . Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$ . Then the previous statement sa... | 113 |
6,351 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_11 | 1 | semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$ | We can just look at a quarter circle inscribed in a $45-45-90$ right triangle. We can then extend a radius, $r$ to one of the sides creating an $r,r, r\sqrt{2}$ right triangle. This means that we have $r + r\sqrt{2} = 8\sqrt{2}$ so $r = \frac{8\sqrt{2}}{1+\sqrt{2}} = 16 - 8\sqrt{2}$ . Then the diameter is $32 - \sqrt{5... | 544 |
6,352 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_11 | 2 | semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$ | We proceed by finding the area of the square in 2 different ways. The square is obviously 8*8=64, but we can also find the area in terms of d. From the center of the circle, draw radii that hit the points where the square is tangent to the semicircle. Then the square's area is the area of the small square +2* the area ... | 544 |
6,353 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_11 | 3 | semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$ | It is easy after getting the image, after drawing labeling the lengths of those segments, assume the radius is $x$ , we can see $x=\sqrt{2}(8-x)$ and we get $2x=32-\sqrt{512}$ and we have the answer $\boxed{544}$ ~bluesoul | 544 |
6,354 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_14 | 1 | Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by $1000$ | Let $(a,b)$ denote a normal vector of the side containing $A$ . Note that $\overline{AC}, \overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$ $ax+by=8a$ $bx-ay=10b-9a$ , and $bx-ay=-4b-7a$ . The lines form a square, so the di... | 936 |
6,355 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_2 | 1 | A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each ... | Use construction . We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined.
Our answer is thus $\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}$ , and $m + n = \boxed{79}$ | 79 |
6,356 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_2 | 2 | A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each ... | Call the three different types of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, and the third three like ABCABCABC or BCABACCAB. This can occur in $\left(\frac{3!}{1!1!1!}\right)^3 = 6^3 = 216$ different manners. The total number of... | 79 |
6,357 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_2 | 3 | A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each ... | The denominator of m/n is equal to the total amount of possible roll configurations given to the three people. This is equal to ${9 \choose 3}{6 \choose3}$ as the amount of ways to select three rolls out of 9 to give to the first person is ${9 \choose 3}$ , and three rolls out of 6 is ${6 \choose3}$ . After that, the t... | 79 |
6,358 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_3 | 1 | An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac mn$ where $m$ and $n$ are relatively prime integers . Find $m+n.$ | Let's call the first term of the original geometric series $a$ and the common ratio $r$ , so $2005 = a + ar + ar^2 + \ldots$ . Using the sum formula for infinite geometric series, we have $\;\;\frac a{1 -r} = 2005$ . Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + \ldots$ . We know this series has sum $20050 =... | 802 |
6,359 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_4 | 1 | Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$ | $10^{10} = 2^{10}\cdot 5^{10}$ so $10^{10}$ has $11\cdot11 = 121$ divisors
$15^7 = 3^7\cdot5^7$ so $15^7$ has $8\cdot8 = 64$ divisors.
$18^{11} = 2^{11}\cdot3^{22}$ so $18^{11}$ has $12\cdot23 = 276$ divisors.
Now, we use the Principle of Inclusion-Exclusion . We have $121 + 64 + 276$ total potential divisors so far, ... | 435 |
6,360 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_4 | 2 | Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$ | We can rewrite the three numbers as $10^{10} = 2^{10}\cdot 5^{10}$ $15^7 = 3^7\cdot5^7$ , and $18^{11} = 2^{11}\cdot3^{22}$ .
Assume that $n$ (a positive integer) is a divisor of one of the numbers. Therefore, $n$ can be expressed as ${p_1}^{e_1}$ or as ${p_2}^{e_2}{p_3}^{e_3}$ where $p_1$ $p_2$ are in $\{2,3,5\}$ and ... | 435 |
6,361 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_5 | 2 | Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$ | Let $k=\log_a b$ . Then our equation becomes $k+\frac{6}{k}=5$ . Multiplying through by $k$ and solving the quadratic gives us $k=2$ or $k=3$ . Hence $a^2=b$ or $a^3=b$
For the first case $a^2=b$ $a$ can range from 2 to 44, a total of 43 values.
For the second case $a^3=b$ $a$ can range from 2 to 12, a total of 11 valu... | 54 |
6,362 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_5 | 3 | Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$ | Using the change of base formula on the second equation to change to base $a$ , we get $\log_a(b) + \frac{6 \log_a(a)}{\log_a(b)}$ . If we substitute $x$ for $\log_a(b)$ , we get $x + \frac{6}{x}$ . Multiplying by $x$ on both sides and solving, we get $x=3,2$ . Substituting back in, we get $\log_a(b) = 3,2$ . That mean... | 54 |
6,363 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_6 | 1 | The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this proc... | Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its original ... | 392 |
6,364 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_6 | 2 | The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this proc... | If you index the final stack $1,2,\dots,2n$ , you notice that pile A resides only in the odd indices and has maintained its original order aside from flipping over. The same has happened to pile B except replace odd with even. Thus, if 131 is still at index 131, an odd number, then 131 must be from pile A. The numbers ... | 392 |
6,365 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_7 | 1 | Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $(x+1)^{48}$ | We note that in general,
It now becomes apparent that if we multiply the numerator and denominator of $\frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) }$ by $(\sqrt[16]{5} - 1)$ , the denominator will telescope to $\sqrt[1]{5} - 1 = 4$ , so
It follows that $(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3... | 125 |
6,366 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_7 | 2 | Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $(x+1)^{48}$ | Like Solution $2$ , let $z=\sqrt[16]{5}$ Then, the expression becomes
$x=\frac{4}{(z+1)(z^2+1)(z^4+1)(z^8+1)}$ Now, multiplying by the conjugate of each binomial in the denominator, we obtain...
$x=\frac{4(z-1)(z^2-1)(z^4-1)(z^8-1)}{(z^2-1)(z^4-1)(z^8-1)(z^{16}-1)}\implies x=\frac{4(z-1)}{z^{16}-1}$ Plugging back in,
$... | 125 |
6,367 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_8 | 1 | Circles $C_1$ and $C_2$ are externally tangent , and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear . A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the cho... | Let $O_1, O_2, O_3$ be the centers and $r_1 = 4, r_2 = 10,r_3 = 14$ the radii of the circles $C_1, C_2, C_3$ . Let $T_1, T_2$ be the points of tangency from the common external tangent of $C_1, C_2$ , respectively, and let the extension of $\overline{T_1T_2}$ intersect the extension of $\overline{O_1O_2}$ at a point $H... | 405 |
6,368 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_8 | 2 | Circles $C_1$ and $C_2$ are externally tangent , and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear . A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the cho... | Call our desired length $x$ . Note for any $X$ on $\overline{AB}$ and $Y$ on $\overline{O_1O_2}$ such that $\overline{XY}\perp\overline{AB}$ that the function $f$ such that $f(\overline{O_1Y})=\overline{XY}$ is linear. Since $(0,4)$ and $(14,10)$ , we can quickly interpolate that $f(10)=\overline{O_3T}=\frac{58}{7}$ . ... | 405 |
6,369 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | 1 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | We know by De Moivre's Theorem that $(\cos t + i \sin t)^n = \cos nt + i \sin nt$ for all real numbers $t$ and all integers $n$ . So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.
Recall the trigonometric identities $\cos \left(\frac{\pi}2 - u\right) = \sin ... | 250 |
6,370 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | 2 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | This problem begs us to use the familiar identity $e^{it} = \cos(t) + i \sin(t)$ . Notice, $\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}$ since $\sin(-t) = -\sin(t)$ . Using this, $(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)$ is recast as $(i e^{-it})^n = i e^{-itn}$ . Hence we must have $i^n = i \Right... | 250 |
6,371 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | 4 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | We are using degrees in this solution instead of radians. I just process stuff better that way.
We can see that the LHS is $cis(n(90^{\circ}-t))$ , and the RHS is $cis(90^{\circ}-nt)$ So, $n(90-t) \equiv 90-nt \mod 360$ Expanding and canceling the nt terms, we will get $90n \equiv 90 \mod 360$ . Canceling gets $n \equi... | 250 |
6,372 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | 5 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | Let $t=0$ . Then, we have $i^n=i$ which means $n\equiv 1\pmod{4}$ . Thus, the answer is $\boxed{250}$ | 250 |
6,373 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | 6 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | We factor out $i^n$ from $(\sin t + i \cos t)^n = i^n (\cos(t) - i \sin t)= i^n(\cos(nt) - i\sin nt).$ We know the final expression must be the same as $\sin nt + i \cos nt$ so we must have $i^n(\cos(nt) - i\sin nt) = \sin nt + i \cos nt$ in which testing yields $n \equiv 1 \pmod{4}$ is the only mod that works, so we h... | 250 |
6,374 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | 7 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | Note that this looks like de Moivre's except switched around. Using de Moivre's as motivation we try to convert the given expression into de Moivre's. Note that $\sin t = \cos(90 - t)$ and $\cos t = \sin(90 - t)$ . So we rewrite the expression and setting it equal to the given expression in the problem, we get $\cos(90... | 250 |
6,375 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | 8 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | $(\sin\theta + i\cos\theta)^{n} = i^{n} (\cos\theta - i\sin\theta)^n$
Hence the required condition is just $i^{n} = i$ which is true for exactly 1 in 4 consecutive numbers. Thus $\boxed{250}$ | 250 |
6,376 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_11 | 1 | Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$ | For $0 < k < m$ , we have
Thus the product $a_{k}a_{k+1}$ is a monovariant : it decreases by 3 each time $k$ increases by 1. For $k = 0$ we have $a_{k}a_{k+1} = 37\cdot 72$ , so when $k = \frac{37 \cdot 72}{3} = 888$ $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m = \boxed{889}$ , our answer. | 889 |
6,377 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_11 | 2 | Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$ | Plugging in $k = m-1$ to the given relation, we get $0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}$ . Inspecting the value of $a_{k}a_{k+1}$ for small values of $k$ , we see that $a_{k}a_{k+1} = 37\cdot 72 - 3k$ . Setting the RHS of this equation equal to $3$ , we find that $m$ must be $\boxed{889}$ | 889 |
6,378 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | 1 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | Let $G$ be the foot of the perpendicular from $O$ to $AB$ . Denote $x = EG$ and $y = FG$ , and $x > y$ (since $AE < BF$ and $AG = BG$ ). Then $\tan \angle EOG = \frac{x}{450}$ , and $\tan \angle FOG = \frac{y}{450}$
By the tangent addition rule $\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)$ ,... | 307 |
6,379 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | 2 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | [asy] size(3inch); pair A, B, C, D, M, O, X, Y; A = (0,900); B = (900,900); C = (900,0); D = (0,0); M = (450,900); O = (450,450); X = (250 - 50*sqrt(7),900); Y = (650 - 50*sqrt(7),900); draw(A--B--C--D--cycle); draw(X--O--Y); draw(M--O--A); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); lab... | 307 |
6,380 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | 3 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with $o = 0, a = -450 + 450i, b = 450 + 450i$ , and $f = x + 450i$ . Since $EF$ = 400, $e = (x-400) + 450i$ . From $\angle{EOF} = 45^{\circ}$ , we can deduce that the rotation of point $F$ 45 degrees cou... | 307 |
6,381 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | 4 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | [asy] size(250); pair A,B,C,D,O,E,F,G,H,K; A = (0,0); B = (900,0); C = (900,900); D = (0,900); O = (450,450); E = (600,0); F = (150,0); G = (-600,0); H = (450,0); K = (0,270); draw(A--B--C--D--cycle); draw(O--E); draw(O--F); draw(O--G); draw(A--G); draw(O--H); label("O",O,N); label("A",A,S); label("B",B,SE); label("C",... | 307 |
6,382 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | 5 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | We know that G is on the perpendicular bisector of $EF$ , which means that $EJ=JF=200$ $EG=GF=200\sqrt{2}$ and $GH=250$ . Now, let $HO$ be equal to $x$ . We can set up an equation with the Pythagorean Theorem:
\begin{align*} \sqrt{x^2+250^2}&=(200\sqrt{2})^2 \\ x^2+62500&=80000 \\ x^2&=17500 \\ x&=50\sqrt{7} \end{align... | 307 |
6,383 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_13 | 1 | Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1\cdot n_2.$ | We define $Q(x)=P(x)-x+7$ , noting that it has roots at $17$ and $24$ . Hence $P(x)-x+7=A(x-17)(x-24)$ . In particular, this means that $P(x)-x-3=A(x-17)(x-24)-10$ . Therefore, $x=n_1,n_2$ satisfy $A(x-17)(x-24)=10$ , where $A$ $(x-17)$ , and $(x-24)$ are integers. This cannot occur if $x\le 17$ or $x\ge 24$ because th... | 418 |
6,384 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_13 | 2 | Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1\cdot n_2.$ | We know that $P(n)-(n+3)=0$ so $P(n)$ has two distinct solutions so $P(x)$ is at least quadratic. Let us first try this problem out as if $P(x)$ is a quadratic polynomial. Thus $P(n)-(n+3)= an^2+(b-1)n+(c-3)=0$ because $P(n)=an^2+bn+c$ where $a,b,c$ are all integers. Thus $P(x)=ax^2+bx+c$ where $a,b,c$ are all integers... | 418 |
6,385 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | 1 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | By the Law of Sines and since $\angle BAE = \angle CAD, \angle BAD = \angle CAE$ , we have
\begin{align*} \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\ &= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\ &= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{B... | 463 |
6,386 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | 2 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively.
From here, we can use Heron's Formula to find the altitude. The area of the triangle is $\sqrt{21*6*7*8} = 84$ . We can then use similar tri... | 463 |
6,387 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | 3 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Let $\angle CAD = \angle BAE = \theta$ . Note by Law of Sines on $\triangle BEA$ we have \[\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}\] As a result, our goal is to find $\sin{\angle BEA}$ and $\sin{\theta}$ (we already know $AB$ ).
Let the foot of the altitude from $A$ to $BC$ be $H$ . B... | 463 |
6,388 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | 4 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Since $AE$ and $AD$ are isogonal with respect to the $A$ angle bisector, we have \[\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.\] To prove this, let $\angle BAE=\angle DAC=x$ and $\angle BAD=\angle CAE=y.$ Then, by the Ratio Lemma, we have \[\frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}\] \[\frac{BE}{EC}=\frac{AB\sin ... | 463 |
6,389 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | 5 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Let $ED = x$ , such that $BE = 9-x$ . Since $\overline{AE}$ and $\overline{AD}$ are isogonal, we get $\frac{9-x}{6+x} \cdot \frac{9}{6} = \frac{13^2}{14^2} \Rightarrow 588(9 - x) = 338(6 + x)$ , and we can solve to get $x = \frac{1632}{463}$ (and $BE = \frac{2535}{463}$ ). Hence, our answer is $\boxed{463}$ . - Spacesa... | 463 |
6,390 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | 6 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Diagram borrowed from Solution 1.
Applying Law of Cosines on $\bigtriangleup ABC$ with respect to $\angle C$ we have \[AB^2=AC^2+BC^2-2(AC)(BC)\cos C\] Solving gets $\cos C=\frac{3}{5}$ , which implies that \[\sin C=\sqrt{1-\cos C}=\frac{4}{5}\] Applying Stewart's Theorem with cevian $AD$ we have \[(BC)(AD)^2+(BC)(BD)(... | 463 |
6,391 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15 | 1 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and ... | Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$
Let $w_3$ have center $(x,y)$ and radius $r$ . Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$ , and if they are internally tangent, it is $|r_1 - r_2|$ ... | 169 |
6,392 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15 | 2 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and ... | As above, we rewrite the equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$ . Let $F_1=(-5,12)$ and $F_2=(5,12)$ . If a circle with center $C=(a,b)$ and radius $r$ is externally tangent to $w_2$ and internally tangent to $w_1$ , then $CF_1=16-r$ and $CF_2=4+r$ . Therefore, $CF_1+CF_2=20$ . In particu... | 169 |
6,393 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15 | 3 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and ... | We use the same reflection as in Solution 2. As $OF_1'=OF_2=13$ , we know that $\triangle OF_1'F_2$ is isosceles. Hence $\angle F_2F_1'O=\angle F_1'F_2O$ . But by symmetry, we also know that $\angle OF_1T=\angle F_2F_1'O$ . Hence $\angle OF_1T=\angle F_1'F_2O$ . In particular, as $\angle OF_1T=\angle OF_2T$ , this impl... | 169 |
6,394 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15 | 4 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and ... | First, rewrite the equations for the circles as $(x+5)^2+(y-12)^2=16^2$ and $(x-5)^2+(y-12)^2=4^2$ .
Then, choose a point $(a,b)$ that is a distance of $x$ from both circles. Use the distance formula between $(a,b)$ and each of $A$ and $C$ (in the diagram above). The distances, as can be seen in the diagram above a... | 169 |
6,395 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_1 | 1 | The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by $37$ | A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form ${\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}}$ $= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n$ , for $n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrac... | 217 |
6,396 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | 1 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | Note that since set $A$ has $m$ consecutive integers that sum to $2m$ , the middle integer (i.e., the median) must be $2$ . Therefore, the largest element in $A$ is $2 + \frac{m-1}{2}$
Further, we see that the median of set $B$ is $0.5$ , which means that the "middle two" integers of set $B$ are $0$ and $1$ . Therefo... | 201 |
6,397 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | 2 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | Let us give the elements of our sets names: $A = \{x, x + 1, x + 2, \ldots, x + m - 1\}$ and $B = \{y, y + 1, \ldots, y + 2m - 1\}$ . So we are given that \[2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,\] so $2 = x + \frac{m - 1}2$ and $x + (m - 1) = \frac{m + 3}2$ (... | 201 |
6,398 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | 3 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | The thing about this problem is, you have some "choices" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.
First, we note that for set $A$
\[\frac{m(f + l)}{2} = 2m\]
Where $f$ and $l$ represent the first and ... | 201 |
6,399 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | 4 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | First, calculate the average of set $A$ and set $B$ . It's obvious that they are $2$ and $1/2$ respectively.
Let's look at both sets. Obviously, there is an odd number of integers in the set with $2$ being in the middle, which means that $m$ is an odd number and that the number of consecutive integers on each side of ... | 201 |
6,400 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | 5 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | Let the first term of $A$ be $a$ and the first term of $B$ be $b$ . There are $m$ elements in $A$ so $A$ is $a, a+1, a+2,...,a+m-1$ . Adding these up, we get $\frac{2a+m-1}{2}\cdot m = 2m \implies 2a+m=5$ . Set $B$ contains the numbers $b, b+1, b+2,...,b+2m-1$ . Summing these up, we get $\frac{2b+2m-1}{2}\cdot 2m =m \i... | 201 |
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