Solution
Square Divisors of a Factorial
Show the problem again
How many positive divisors of $12! = 479001600$ are perfect squares? (Count 1 as a perfect square.) Give an integer.
Worked solution
The answer is 36. By Legendre's formula, $12! = 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11$. A square divisor $d^2$ needs every prime exponent even and at most the exponent in $12!$. The choices are: exponent of 2 in {0, 2, 4, 6, 8, 10} (6 ways), of 3 in {0, 2, 4} (3 ways), of 5 in {0, 2} (2 ways), and 0 for both 7 and 11. That gives 6 × 3 × 2 = 36 square divisors.
Source: Classic exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.