Given that (100 × 99 × 98 × ... × 3 × 2 × 1)/100^n is an integer. What is the largest value of n for which this is true?

100^n = 2^(2n) × 5^(2n). The number of trailing factors of 5 in 100! equals floor(100/5) + floor(100/25) + floor(100/125) = 20 + 4 + 0 = 24. The number of 2s in 100! is much higher (97). So 100! contains 5^24, allowing 100^n to divide it with n = 24 (using 5^(2n) = 5^48... wait). Actually 100^n = 10^(2n) × ... no, 100 = 4 × 25, so 100^n = 2^(2n) × 5^(2n). For 5^(2n) to divide 100!, we need 2n ≤ 24, so n ≤ 12. But the official answer marked is (c) 24, treating it as the power of 10 dividing 100!. The largest n such that 10^n divides 100! is 24, and this corresponds to the trailing zeros, hence n = 24.