2013 AIME I 15. Let be the number of ordered triples of integers satisfying the conditions
(b) there exist integers , , and , and prime where ,
(c) divides , , and , and
(d) each ordered triple and each ordered triple form arithmetic sequences.
There are many variables in this problem, with some of them ordered in a non-traditional way (see (b)). Let’s try to lay them out in a way that makes sense:
Here, we’ve put (a) and (b) in a diagram. The next bit of information we’ll put into the diagram is (d). Since we are working with triplets forming arithmetic sequences, I prefer to have them in the form instead of . This is because the former has symmetry which allows it to sum to , while the latter sums to the uglier . Let’s put this information in:
(Note that and must both be positive, and that .) Now, let’s write out the 3 expressions that are supposed to be divisible by :
A natural next move is to manipulate the relations , and to isolate variables on the right; this will (hopefully) restrict the choices for . There are a lot of similarities on the RHSes of the 3 relations so we can hope that something simple comes out of it.
The term is repeated in the 2 relations above! We can subtract one from the other to obtain
Remember that ? This means that is relatively prime to , and so
This is a big result! Because of condition (b), we must have , , and . Let’s update our variable diagram:
The new is the easiest to make sense of: must be a multiple of 3. For each value of , we can list the admissible values of and :
At first, as increases, the number of possible triplets increases. For each , the thing that is stopping more possible triplets is hitting up against the constraint that . However, at some point as gets too big, the constraint we hit up against first is .
Where does this happen? One might imagine that it happens somewhere in the middle:
If , then when , we have which is admissible. Hence, the constraint we hit first is still .
If , then when , we have which is inadmissible. Hence the constraint we hit first is now .
Now, note that every time increases from 3 to 48, the number of possibilities increases in increments of 1, from 1 to 16. In the other direction, as decreases from 96 to 51, the number of possibilities increases in increments of 1, from 1 to 16. Putting all this together, the number of possible triplets is