2013 AIME I Problems/Problem 15
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, .
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
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