2013 AIME I Problems/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 .
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, .
Let = and . Now the 3 differences would be
Adding equations and would give . Then doubling equation would give . The difference between them would be . Since , then . Since is prime, or . However, since , we must have , which means .
If , the only possible values of are . Plugging this into our differences, we get The difference between and is , which should be divisible by 3. So . Also note that since , . Now we can try different values of and :
When , triples.
When , triples..
... and so on until
When , triple.
So the answer is
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