If 10^m × 1000 × n = 75^25 × 25^32 × 32^75, where n is not divisible by 10, then the value of m is

Factorize RHS: 75^25 = (3·5²)^25 = 3^25 · 5^50; 25^32 = 5^64; 32^75 = 2^375. So RHS = 3^25 · 5^(50+64) · 2^375 = 3^25 · 5^114 · 2^375. LHS = 10^m · 1000 · n = 2^m · 5^m · 2^3 · 5^3 · n = 2^(m+3) · 5^(m+3) · n. Since n is not divisible by 10, n contributes no extra pair of 2 and 5. The number of 10s (pairs of 2 and 5) extracted is limited by min(power of 2, power of 5) = min(375, 114) = 114. So m+3 = 114, giving m = 111. Wait, recheck: 10^m gives m twos and m fives, plus 1000 gives 3 twos and 3 fives. So 2^(m+3) and 5^(m+3) must be factors. We need m+3 ≤ 114 (limited by 5s) and remaining factors absorbed in n. For n not divisible by 10, m+3 must equal 114, so m=111. The remaining n = 2^(375-114) · 3^25 = 2^261 · 3^25, which has no factor of 5, hence not divisible by 10.