2013 AIME I Problems/Problem 15
Contents
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 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 . So we have , meaning . Condition implies that or . Now we return to condition , which now implies that . Now, we set for increasing positive integer values of . yields no solutions. gives , giving us solution. If , we get solutions, and . Proceeding in the manner, we see that if , we get 16 solutions. However, still gives solutions because . Likewise, gives solutions. This continues until gives one solution. gives no solution. Thus, .
Solution 2
Condition c gives us that , etc. Condition d then tells us that C and c can be expressed as and , respectively. However, plugging what we got from condition c into this, we find that . From there, we branch off into two cases; either , or . Realize then that the second case leads to a contradiction, due to condition b. Then, means that must be . The bash from there is pretty similar to what was done in Solution 1. We get . - Spacesam
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
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