Problem 7 · Parity of prime factors in divisor counts · 5 points
Made by AI and not reviewed by a real person.
Let N = 2⁵·3⁵·5⁵·7⁵. How many positive divisors d of N have the property that both d and N/d have a perfect-square number of positive divisors?
- 90
- 96
- 120
- 36
- 126
Show a hint
- A number is a square when every prime in its factorization occurs an even number of times. Think about both divisor counts at once.
- For each prime with exponent a in d, its contributions to the two divisor counts are a+1 and 6−a.
Show the answer and solution
Answer: B — 96
For j = 1, 2, …, 6, let n_j count the primes whose exponent in d is j−1. Each contributes j to the divisor count of d and 7−j to that of N/d. For both counts to be squares, the prime factor 5 forces n₅ and n₂ to be even; the factor 2 then forces n₆ and n₁ to be even; and the factor 3 forces n₃ and n₄ to be even. Conversely, these conditions make both counts squares. Since the six counts n_j add to 4, either one j occurs four times (6 ways), or two different j values occur twice each (C(6,2)·C(4,2) = 90 ways). The answer is 96; the tempting 90 forgets the all-four-equal cases.