How many factors of the number 28 * 36 * 54 * 105 are multiples of 120?

Concept: A divisor of N is a multiple of M exactly when, prime by prime, the divisor's exponent is at least M's exponent for every prime that appears in M; for a prime absent from M, the divisor's exponent may range freely from 0 up to N's own exponent for that prime. Each prime therefore contributes an independent count of valid exponents, and the total number of qualifying divisors is the product of these per-prime counts.

Application — express N in a matching prime base:

105 = (2·5)5 = 25 · 55, so combining exponents of 2 gives 8 + 5 = 13, and combining exponents of 5 gives 4 + 5 = 9. Hence N = 213 · 36 · 59.

120 = 23 · 31 · 51 — so a divisor of N that is a multiple of 120 must carry at least these exponents for 2, 3, and 5.

• Exponent of 2: must lie between 3 and 13, inclusive → 13 − 3 + 1 = 11 valid values.

• Exponent of 3: must lie between 1 and 6, inclusive → 6 − 1 + 1 = 6 valid values.

• Exponent of 5: must lie between 1 and 9, inclusive → 9 − 1 + 1 = 9 valid values.

Multiply the counts: 11 × 6 × 9 = 594.

Cross-check: Since 120 divides N, every divisor of N that is a multiple of 120 corresponds bijectively to a divisor of N ÷ 120 = 210 · 35 · 58. The total number of divisors of that number is (10+1)(5+1)(8+1) = 11 × 6 × 9 = 594 — matching the count above.

Conclusion: 594 of the factors of 28 · 36 · 54 · 105 are multiples of 120.