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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