How many perfect square factors does $20^{20} $ have?
I found that $20^{20} = 5^{20}. 2^{40}$.
$5^{2}, 5^{4}, 5^{6}, ... , 5^{20}$ (10 perfect square factors)
$2^{2}, 2^{4}, 2^{6}, ... , 2^{40}$ (20 perfect square factors)
$5^{2}.2^{2}, 5^{2}.2^{4}, ..., 5^{2}.2^{40}$ and like this, there are 20 more perfect square factors for every 10 perfect square factors. So, there are 20 * 10 = 200 factors. Total number of perfect square factor is 200 + 20 +10 = 230.
My question is am I missing something and is there any easy or more generalized way to solve this math?
$\endgroup$ 32 Answers
$\begingroup$If $d \mid 20^{10}$ then $d^2 \mid 20^{20}$ and vice versa. So all you need to do is count the number of divisors of $20^{10}.$ And we have formulas for that:
$$\tau(20^{10}) = \tau(2^{20}5^{10}) = \tau(2^{20})\tau(5^{10}) = (20+1)(10+1) = 231.$$
$\endgroup$ $\begingroup$In general, suppose you had a prime factorization of the form $$N=\prod p_i^{a_i}$$
Then the greatest even power of $p_i$ which divides $N$ has exponent $2\times\big \lfloor \frac {a_i}2\big \rfloor$ so the answer would be $$\prod \left( \Big \lfloor \frac {a_i}2\Big \rfloor+1\right)$$
$\endgroup$
