Mock AIME 2 2006-2007 Problems/Problem 8
This solution is rather long and unpleasant, so a nicer solution may exist:
From the givens, and so and .
Note that this factorization of 144 contains a pair of consecutive integers, and . The factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72 and 144 itself. As both and are positive integers, , so we must have equal to one of 2, 3 and 8.
If then and so from which . It is clear that this equation has no solutions if , and neither nor is a solution, so in this case we have no solutions.
If then so . It is clear that is the unique solution to this equation in positive integers. Then and our sequence is .
If then either:
a) and so so , which has no solutions in positive integers
b) and so so which has solution . Then our sequence becomes .
Thus we see there are two possible sequences, but in both cases the answer is .
A Second Simpler Solution:
We can use smart "guess-and-check" for this problem, seeing as there are not that many options anyways. We know that we need factors of to be and We can also infer that will likely need to be one of the smaller factors.
The factors of are:
Selecting numbers from this list and trying them out, we can satisfy the conditions with these numbers:
Therefore, the answer is
|Mock AIME 2 2006-2007 (Problems, Source)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|