NIMCET 2009 — Reasoning PYQ

To find the remainder of the sum $x={\sum }_{n=1}^{100}n!$ when divided by $240$, we need to find the point in the series where the terms become divisible by $240$. Once a term $n!$ is divisible by $240$, all subsequent terms $(n+1)!,(n+2)!,\dots$ will also be divisible by $240$ because they contain $n!$ as a factor.

Step 1: Find the prime factorization of 240.

Step 2: Determine which $n!$ is first divisible by 240.

Let's check the values of factorials:

• $1!=1$

• $2!=2$

• $3!=6$

• $4!=24$

• $5!=120$

• $6!=720$

Since $720=240\times 3$, we see that $6!$ is exactly divisible by $240$.

Therefore, for all $n\ge 6$, the remainder of $n!$ when divided by $240$ is $0$.

Step 3: Calculate the sum of the terms before $6!$.

We only need to find the remainder of the sum of the first five terms:

Step 4: Find the final remainder.

Since the sum of the first five terms is $153$ and all subsequent terms from $6!$ to $100!$ have a remainder of $0$ when divided by $240$:

The remainder is $153$.

Correct Option:

(a) 153