Let n = p2q, where p and q are distinct prime numbers. How many numbers m…

Concept. The count of integers m with 1 ≤ m ≤ n and gcd(m, n) = 1 is Euler's totient φ(n). Two governing properties: for a prime power pk, φ(pk) = pk − pk−1 (remove the multiples of p), and φ is multiplicative on coprime factors, φ(ab) = φ(a)·φ(b) when gcd(a, b) = 1.

Application.

1. Factor n = p2q with p, q distinct primes, so p2 and q are coprime and φ(p2q) = φ(p2)·φ(q).

2. φ(p2): among 1, …, p2 the values sharing a factor with p2 are exactly the multiples of p, and there are p of them, so φ(p2) = p2 − p = p(p − 1).

3. φ(q): for a prime q every smaller positive integer is coprime to it, so φ(q) = q − 1.

4. Multiply: φ(p2q) = p(p − 1)(q − 1).

Cross-check. Take p = 3, q = 5, so n = p2q = 45. The formula gives p(p − 1)(q − 1) = 3·2·4 = 24, and a direct count finds exactly 24 integers in 1…45 that are coprime to 45 — and this value differs from every other option (12, 15, 32), confirming the result.

So the number of such m is p(p − 1)(q − 1).