These Divisors Are Too Large!

$\text {For prime numbers, the number of divisors, } \phi (p^n) = n + 1$ .

For composite numbers:

$c = (p_1) ^{m_1} × (p_2) ^{m_2} × ... × (p_k) ^{m_k}$

$\text {where } p_1, p_2, ..., p_k \text { are distinct prime numbers and } k, m_1, m_2, ..., m_k \in \mathbb {N}$

$\phi (c^n) = (nm_1 + 1) × (nm_2 + 1) × ... × (nm_k + 1)$

In this case, 31, 41 and 61 are prime numbers, so their nth power has exactly (n + 1) divisors.

51 is a composite number, since it's prime factorisation is:

$3^1 × 17^1$

Therefore ( c = n = 51),

$\phi (51^n) = (n × 1 + 1) × (n × 1 + 1) = (n + 1)^2 > n +1$

$52^2 > 52$

Hence, our answer should be:

$\boxed {51}$

Normally, a prime number raised to a power n would have divisors as that same number raised to the power of n-1, n-2,...,1. Therefore option A does not follow suit as it is not a prime number.

51 is not prime.

1,(n^2),...,(n^n) are the obvious (n+1)divisors of n^n.The no of divisors is greater if n is composite, and it is these n+1 numbers only if n is a prime.