NIMCET 2021 — Reasoning PYQ

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Question

NIMCET 2021 — Reasoning PYQ

NIMCET | Reasoning | 2021

Choose the correct option for the remainder when X=1!+2!+3!+…+100! is divided by 24.

Choose the correct answer:

Explanation

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:

$n! \equiv 0 \pmod{24} \quad \text{for all } n \ge 4$

3. Simplify the expression for the remainder:

The original sum is:

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

Taking the modulo $24$ of the entire sum:

$X \equiv (1! + 2! + 3!) + (4! + 5! + \dots + 100!) \pmod{24}$

$X \equiv (1! + 2! + 3!) + (0 + 0 + \dots + 0) \pmod{24}$

4. Perform final calculation:

$X \equiv 1 + 2 + 6 \pmod{24}$

$X \equiv 9 \pmod{24}$

Result:

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

Correct Option:

$9$