If N = 23 × 34 × 52, find the total number of even factors of N.

Concept: If a number N has prime factorization N = (prime)exponent repeated over each of its prime factors, the total number of factors of N is the product of (exponent + 1) taken over all its primes. A factor of N is odd exactly when it excludes the prime 2 entirely, so the count of odd factors uses the same product formula but with the prime-2 term dropped. Even factors of N = Total factors − Odd factors.

1. Prime-factorize N: N = 23·34·52 — exponents 3, 4, and 2 on the primes 2, 3, and 5.

2. Total factors of N = (3+1)(4+1)(2+1) = 4 × 5 × 3 = 60.

3. Odd factors of N (exponent of 2 fixed at 0) = (4+1)(2+1) = 5 × 3 = 15.

4. Even factors of N = Total factors − Odd factors = 60 − 15 = 45.

Cross-check (direct count): The exponent of 2 in an even factor must be 1, 2, or 3 (3 choices, excluding 0); the exponent of 3 can independently be any of 0–4 (5 choices); the exponent of 5 can be any of 0–2 (3 choices). Multiplying the independent choices gives 3 × 5 × 3 = 45, matching the subtraction method above.

Answer: 45