# 2019 AIME I Problems/Problem 9

## Problem 9

Let $\tau (n)$ denote the number of positive integer divisors of $n$. Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$.

## Solution

Essentially, you realize that to get 7 you need an odd amount of divisors + and even amount of divisors. This means that one of our n needs to be a square. Furthermore it must either be a prime squared to get 3 divisors or a prime to the fourth to get 5 divisors. Any more factors in a square would be to large. Thus n/n+1 is in the form p^2 or p^4. The rest of the solution is bashing left to the reader.

~~ paliwalar.21

## See Also

 2019 AIME I (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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