# 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

In order to obtain a sum of 7, we must have:

• either a number with 5 divisors (a fourth power of a prime) and a number with 2 divisors (a prime), or
• a number with 4 divisors (a semiprime) and a number with 3 divisors (a square of a prime). (No integer greater than 1 can have fewer than 2 divisors.)

Since both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square like $3^2$ with 3 divisors, or a fourth power like $2^4$ with 5 divisors. We then find the smallest such values by hand.

• $2^2$ has two possibilities: 3 and 4, or 4 and 5. Neither works.
• $3^2$ has two possibilities: 8 and 9, or 9 and 10. (8,9) and (9,10) both work.
• $2^4$ has two possibilities: 15 and 16, or 16 and 17. Only (16,17) works.
• $5^2$ has two possibilities: 24 and 25, or 25 and 26. Only (25,26) works.
• $7^2$ has two possibilities: 48 and 49, or 49 and 50. Neither works.
• $3^4$ has two possibilities: 80 and 81, or 81 and 82. Neither works.
• $11^2$ has two possibilities: 120 and 121, or 121 and 122. Only (121,122) works.
• $13^2$ has two possibilities: 168 and 169, or 169 and 170. Neither works.
• $17^2$ has two possibilities: 288 and 289, or 289 and 290. Neither works.
• $19^2$ has two possibilities: 360 and 361, or 361 and 362. Only (361,362) works.

Having computed the working possibilities, we take the sum of the corresponding values of $n$: $8+9+16+25+121+361 = \boxed{540}$. ~Kepy.

Possible improvement: since all primes $>2$ are odd, their fourth powers are odd as well, which cannot be adjacent to any primes because both of the adjacent numbers will be even. Thus, we only need to check 16 for the fourth power case. - mathleticguyyy

## 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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