Explanation

Trailing zeros equal min(v₂, v₅) of the product. The exponent of 5 is Σ v₅(k)·(51−k) over multiples of 5: k = 5, 10, 15, 20 give 46+41+36+31; k = 25 (5²) gives 2×26 = 52; k = 30, 35, 40, 45 give 21+16+11+6; k = 50 = 2×5² gives 2×1 = 2. The total is 262, while v₂ = 1139, so the minimum is 262. Hence there are 262 zeros, option (a). The common error is counting only one factor of 5 in 50.