Let *n* be a positive integer with exactly 2 positive prime divisors. If *n*^{2} has 27 divisors, how many does *n* have?

Vanilla Feb 23, 2021

#1**+1 **

Let the 2 positive primes be a and b. Then we have n = a^c * b^c. And if we square this, we see that n^2 = a^(2c) * b ^ (2d). So the number of divisors n^2 has is (2c+1)(2d+1).

Let's take 12 for example.

12 = 2^2 * 3. So we have 3 options for taking 2, 2^0, 2^1, 2^2. Similarly, we have 2 options for taking 3, 3^0, and 3^1. We can see that if x is the exponent, then a^x = n, then n will have x+1 divisors. Back to the 12 examples. Since we have 3 options and 2 options for 2 and 3 respectively, the number of divisors 12 has is 6.

So yeah, n^2 has (2c+1)(2d+1) divisors. We know this is equal to 27. So we have (2c + 1)(2d + 1) = 27. No need to expand the left side, we know the factors of 27 are 1, 3, 9, and 27. So we can test each pair, (1, 27) and (3, 9). Go with the first pair. We have 2c + 1 = 1, meaning c = 0. This logically doesn't make sense, because the problem says "exactly 2 positive prime divisors". So then we know the second pair must work. We plug the second pair in, to get 2c + 1 = 3, meaning 2c = 4 meaning c = 2. 2d + 1 = 9 meaning 2d = 8 meaning d = 4. So c = 2 and d = 4. From this, we apply our little formula to n, getting (c + 1)(d + 1) = number of divisors, so (2 + 1) * (4 + 1) = 3 * 5 = 15.

Note: I intentionally made a mistake above, to prevent CHEATING and just saying the answer is 15 without actually knowing why the answer is 15. Plus, 15 is the wrong answer. Reply back to me what YOU think is the right answer.

Also, someone edit this so there is latex. I suck at latex so........

:P

Zekken Feb 23, 2021