Animated Solution for Mathematics - Permutations and Combinations: The largest value of $n$, for which $40^n$ divides $60!$, is

Imagine you are standing before a colossal mountain of numbers: $60!$. This is the product of every integer from $1$ to $60$, a value so vast it dwarfs the number of stars in our galaxy.

Our mission is to find the largest integer $n$ such that $40^n$ divides this behemoth. In the world of JEE Advanced, we do not brute-force; we decompose the problem into its fundamental prime components.

$40$ is a composite number masking its true prime identity. To reveal its structure, we perform prime factorization:

If we want to form $40^n$, we are effectively looking for $n$ copies of $2^3$ and $n$ copies of $5$. Mathematically, this is expressed as:

This transformation changes the game entirely. We no longer care about the number $40$; we care about how many $2$s and $5$s are lurking inside $60!$. We need $3n$ twos and $n$ fives, and we must determine which resource runs out first.

To count these primes, we use Legendre's Formula. It allows us to calculate the exponent of a prime $p$ in $m!$ without calculating the factorial itself:

First, we calculate the exponent of $5$ in $60!$:

We have exactly $14$ fives available. This implies that our $n$ cannot exceed $14$, as we would run out of fives to complete the factor of $40$.

Next, we calculate the exponent of $2$ in $60!$:

Performing the arithmetic, we find:

We have $56$ twos available. Since each $40$ requires three twos, our constraint for $n$ is:

Since $n$ must be an integer, the constraint from the twos is $n\le 18$.

We now compare our two conditions:

1. From the fives: $n\le 14$ 2. From the twos: $n\le 18$

For $40^n$ to divide $60!$, both conditions must be satisfied simultaneously. The 'binding constraint' is the smaller of the two, as it limits the total number of $40$s we can construct.

Thus, the largest integer $n$ is $14$.