# Difference between revisions of "2013 AIME I Problems/Problem 15"

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==Problem 15== | ==Problem 15== | ||

− | Let <math>N</math> be the number of ordered | + | Let <math>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions (a) <math>0\le A<B<C\le99</math>, (b) there exist integers <math>a</math>, <math>b</math>, and <math>c</math>, and prime <math>p</math> where <math>0\le b<a<c<p</math>, (c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and (d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</math>. |

==Solution== | ==Solution== | ||

From condition (d), we have <math>(A,B,C)=(B-\Delta,B,B+\Delta)</math> and <math>(b,a,c)=(a-\delta,a,a+\delta)</math>. Condition (c) states that <math>p|B-\Delta-a</math>, <math>p|B-a+\delta</math>, and <math>p|B+\Delta-a-\delta</math>. We subtract the first two to get <math>p|-\delta-\Delta</math>, and we do the same for the last two to get <math>p|2\delta-\Delta</math>. We subtract these two to get <math>p|3\delta</math>. So <math>p|3</math> or <math>p|\delta</math>. The second case is clearly impossible, because that would make <math>c=a+\delta>p</math>, violating condition (b). So we have <math>p|3</math>, meaning <math>p=3</math>. Condition (b) implies that <math>(b,a,c)=(0,1,2)</math>. Now we return to condition (c), which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us one solution. If <math>B=6</math>, we get two solutions. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives 16 solutions. <math>B=54</math> gives 15 solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=\boxed{272}</math>. | From condition (d), we have <math>(A,B,C)=(B-\Delta,B,B+\Delta)</math> and <math>(b,a,c)=(a-\delta,a,a+\delta)</math>. Condition (c) states that <math>p|B-\Delta-a</math>, <math>p|B-a+\delta</math>, and <math>p|B+\Delta-a-\delta</math>. We subtract the first two to get <math>p|-\delta-\Delta</math>, and we do the same for the last two to get <math>p|2\delta-\Delta</math>. We subtract these two to get <math>p|3\delta</math>. So <math>p|3</math> or <math>p|\delta</math>. The second case is clearly impossible, because that would make <math>c=a+\delta>p</math>, violating condition (b). So we have <math>p|3</math>, meaning <math>p=3</math>. Condition (b) implies that <math>(b,a,c)=(0,1,2)</math>. Now we return to condition (c), which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us one solution. If <math>B=6</math>, we get two solutions. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives 16 solutions. <math>B=54</math> gives 15 solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=\boxed{272}</math>. |

## Revision as of 21:21, 15 March 2013

## Problem 15

Let be the number of ordered triples of integers satisfying the conditions (a) , (b) there exist integers , , and , and prime where , (c) divides , , and , and (d) each ordered triple and each ordered triple form arithmetic sequences. Find .

## Solution

From condition (d), we have and . Condition (c) states that , , and . We subtract the first two to get , and we do the same for the last two to get . We subtract these two to get . So or . The second case is clearly impossible, because that would make , violating condition (b). So we have , meaning . Condition (b) implies that . Now we return to condition (c), which now implies that . Now, we set for increasing integer values of . yields no solutions. gives , giving us one solution. If , we get two solutions. Proceeding in the manner, we see that if , we get 16 solutions. However, still gives 16 solutions. gives 15 solutions. This continues until gives one solution. gives no solution. Thus, .