Q28 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
The largest value of $n$, for which $40^{n}$ divides $60!$, is
- A11
- B12
- Correct answerC14
- D13
Solution
$40^n = 2^{3n} \times 5^n$. We need the highest power of $40$ dividing $60!$.
Power of $5$ in $60! = \lfloor 60/5 \rfloor + \lfloor 60/25 \rfloor = 12 + 2 = 14$.
Power of $2$ in $60! = \lfloor 60/2 \rfloor + \lfloor 60/4 \rfloor + \lfloor 60/8 \rfloor + \lfloor 60/16 \rfloor + \lfloor 60/32 \rfloor = 30 + 15 + 7 + 3 + 1 = 56$.
We need $n \leq 14$ and $3n \leq 56$ (i.e., $n \leq 18$).
The binding constraint is $n \leq 14$, so the largest value is $n = 14$.
Power of $5$ in $60! = \lfloor 60/5 \rfloor + \lfloor 60/25 \rfloor = 12 + 2 = 14$.
Power of $2$ in $60! = \lfloor 60/2 \rfloor + \lfloor 60/4 \rfloor + \lfloor 60/8 \rfloor + \lfloor 60/16 \rfloor + \lfloor 60/32 \rfloor = 30 + 15 + 7 + 3 + 1 = 56$.
We need $n \leq 14$ and $3n \leq 56$ (i.e., $n \leq 18$).
The binding constraint is $n \leq 14$, so the largest value is $n = 14$.
Hence, the correct answer is option: 'C'.