What is the largest value for n (natural number) such that 6 divides the product of the first 100 natural numbers?

To find the largest value of n such that 6^n divides the product of the first 100 natural numbers (100!), we must determine the exponent of 6 in the prime factorization of 100!. Since 6 is a composite number (2 × 3), its exponent is determined by the minimum of the exponents of its prime factors, 2 and 3. The exponent of a prime p in n! is calculated using Legendre's formula: ⌈n/p⌉ + ⌈n/p²⌉ +.... For prime 3, the calculation is ⌈100/3⌉ + ⌈100/9⌉ + ⌈100/27⌉ + ⌈100/81⌉, which equals 33 + 11 + 3 + 1 = 48. For prime 2, the exponent is significantly higher (97). Since 6 requires one 2 and one 3, the limiting factor is the number of 3s. Thus, the largest value for n is 48. The option '4S' is a typographical representation of 48.