NIMCET 2021 — Reasoning PYQ

1. Calculate initial factorials:

• $1!=1$

• $2!=2$

• $3!=6$

• $4!=24$

• $5!=120$

2. Analyze divisibility by $24$:

We know that for any integer $n\ge 4$, the factorial $n!$ will have $4!$ as a factor.

Since $4!=24$, any $n!$ where $n\ge 4$ is exactly divisible by $24$.

Mathematically:

$n!\equiv 0(\mathrm{mod}24)\text{for all }n\ge 4$

3. Simplify the expression for the remainder:

The original sum is:

$X=1!+2!+3!+4!+5!+\cdots +100!$

Taking the modulo $24$ of the entire sum:

$X\equiv (1!+2!+3!)+(4!+5!+\cdots +100!)(\mathrm{mod}24)$

$X\equiv (1!+2!+3!)+(0+0+\cdots +0)(\mathrm{mod}24)$

4. Perform final calculation:

$X\equiv 1+2+6(\mathrm{mod}24)$

$X\equiv 9(\mathrm{mod}24)$

Result:

The remainder when $X$ is divided by $24$ is $9$.

Correct Option:

$9$