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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