The problem asks for the largest natural number $n$ such that $6^n$ divides the product of the first 100 natural numbers, which is $100!$.

Prime Factorization of Divisor

First, find the prime factorization of the base, 6:

$6 = 2 \times 3$

Therefore, $6^n = (2 \times 3)^n = 2^n \times 3^n$. For $6^n$ to divide $100!$, both $2^n$ and $3^n$ must divide $100!$. This means $n$ must be less than or equal to the exponent of 2 in the prime factorization of $100!$ and also less than or equal to the exponent of 3 in the prime factorization of $100!$. The largest possible value for $n$ will be determined by the minimum of these two exponents.

Exponent Calculation using Legendre's Formula

We use Legendre's formula to find the exponent of a prime $p$ in the factorization of $m!$, which is given by:

$E_p(m!) = \sum_{k=1}^{\infty} \lfloor \frac{m}{p^k} \rfloor$

We need to calculate the exponents for primes 2 and 3 in $100!$.

Exponent of Prime 3 in 100!

Calculate the exponent of 3:

$E_3(100!) = \lfloor \frac{100}{3} \rfloor + \lfloor \frac{100}{3^2} \rfloor + \lfloor \frac{100}{3^3} \rfloor + \lfloor \frac{100}{3^4} \rfloor + \dots$

$E_3(100!) = \lfloor \frac{100}{3} \rfloor + \lfloor \frac{100}{9} \rfloor + \lfloor \frac{100}{27} \rfloor + \lfloor \frac{100}{81} \rfloor$

$E_3(100!) = 33 + 11 + 3 + 1$

$E_3(100!) = 48$

Exponent of Prime 2 in 100!

Calculate the exponent of 2:

$E_2(100!) = \lfloor \frac{100}{2} \rfloor + \lfloor \frac{100}{4} \rfloor + \lfloor \frac{100}{8} \rfloor + \lfloor \frac{100}{16} \rfloor + \lfloor \frac{100}{32} \rfloor + \lfloor \frac{100}{64} \rfloor + \dots$

$E_2(100!) = 50 + 25 + 12 + 6 + 3 + 1$

$E_2(100!) = 97$

Determining the Largest Value of n

The exponent $n$ must satisfy both $n \le E_2(100!)$ and $n \le E_3(100!)$.

$n \le \min(E_2(100!), E_3(100!))$

$n \le \min(97, 48)$

$n \le 48$

The largest natural number $n$ is 48.