How many positive divisors does 10! have, how many of them are perfect squares, and how many are divisible by 100?

How many positive divisors does 10! have, how many of them are perfect squares, and how many are divisible by 100?

Approach: Factor the factorial into primes by counting multiples of each prime and its powers, then translate each of the three questions into a count of admissible exponent vectors.

270, 30 and 70. Legendre's formula counts the exponent of a prime p in 10! as the sum of floor(10/p^k). For 2 that is 5 + 2 + 1 = 8, for 3 it is 3 + 1 = 4, for 5 it is 2, and for 7 it is 1, so the prime factorisation is 10! = 2^8 * 3^4 * 5^2 * 7. Divisor counting multiplies one more than each exponent, giving 9 * 5 * 3 * 2 = 270. A divisor is a perfect square when every entry of its exponent vector is even, leaving 5 choices for the power of 2, 3 for 3, 2 for 5 and 1 for 7, so 5 * 3 * 2 * 1 = 30. Divisibility by 100 = 2^2 * 5^2 requires the exponent of 2 to be at least 2, leaving 7 choices, and the exponent of 5 to be exactly 2, leaving 1 choice, so the count is 7 * 5 * 1 * 2 = 70.

Follow-up: What is the product of all 270 divisors of 10!, expressed as a power of 10!, and why does the parity of the divisor count matter?

Key concepts: prime factorisation, divisor counting, exponent vector, legendre formula.