JEE Mathematics 2026 Question with Solution

Given: We need the largest value of $n$ such that $40^n$ divides $60!$.

Find: The maximum integer $n$.

Prime factorize the base:

So,

Now find the powers of $2$ and $5$ in $60!$ using Legendre's formula.

Power of $2$ in $60!$:

Power of $5$ in $60!$:

For $40^n = 2^{3n} \times 5^n$ to divide $60!$, both of the following must hold:

and

From the first condition,

Therefore the limiting condition is $n \le 14$, so the largest possible value is $14$.

The correct option is C.

Given: $40 = 2^3 \times 5$ and we want the largest $n$ such that $40^n$ divides $60!$.

Find: The maximum $n$.

Compare required prime powers directly:

• each factor of $40^n$ needs $3n$ powers of $2$
• and $n$ powers of $5$

In $60!$, the power of $2$ is $56$ and the power of $5$ is $14$.

So,

Hence,

Therefore, the largest value is $14$, so the correct option is C.