Question

The remainder when $1! + 2! + 3! + .... + 100!$ is divided by $240$ is
A) $153$
B) $33$
C) $73$
D) $187$

Answer 1

Hint: In this question, we have to use the concept of factorials. Factorial of a number $n$ is written as $n!$ . A factorial function multiplies a number by each number below it until $1$ . For example, $4! = 4 \times 3 \times 2 \times 1 = 24$ . This way we will find the terms in \[1! + 2! + 3! + .... + 100!\] that are divisible by $240$ and the terms that are not.

Complete step by step answer:
We have to find the remainder when $1! + 2! + 3! + .... + 100!$ is divided by $240$
$\dfrac{{1! + 2! + 3! + .... + 100!}}{{240}}$ …(1)
On prime factorization of $240$ , we see that:
$240 = 2 \times 2 \times 2 \times 2 \times 3 \times 5$
On rearranging the terms on the right hand side and simplifying, we get:
$240 = 2 \times 5 \times 4 \times 3 \times 2$
Now, multiply and divide the right hand side by $6$
$
   \Rightarrow 240 = \dfrac{{2 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{6} \\
   \Rightarrow 240 = \dfrac{{6!}}{3} \\
 $
When we put this value of $240$ in (1), we get:
$
  \dfrac{{1! + 2! + 3! + .... + 100!}}{{240}} = \dfrac{{1! + 2! + 3! + 4! + 5! + 6! + 7! + 8!.... + 100!}}{{\dfrac{{6!}}{3}}} \\
   \Rightarrow \dfrac{{1! + 2! + 3! + .... + 100!}}{{240}} = \dfrac{{3(1! + 2! + 3! + 4! + 5! + 6! + 7 \times 6! + 8 \times 7 \times 6!.... + 100!)}}{{6!}} \\
 $
In the above obtained equation, we see that every term after $6!$ can be written as a product of $6!$ and other consecutive constants, we don’t expand $100!$ as the expansion would be too big.
Further, we see that the terms before $6!$ do not contain $6!$ in their expansion, so we will separate the two series as follows:
$\dfrac{{1! + 2! + 3! + .... + 100!}}{{240}} = 3(\dfrac{{1! + 2! + 3! + 4! + 5!}}{{6!}}) + 6!(\dfrac{{1 + 7 + 8 \times 7 + ...}}{{6!}})$
Thus, we see that the second term in the right hand side of the above equation is divisible by $6!$ while the first term is not. So, $1! + 2! + 3! + 4! + 5!$ is the remainder when $1! + 2! + 3! + .... + 100!$ is divided by $240$ .
Thus, Remainder $ = 153$
The correct option is option A.

Note: We can also solve the question by simply using the information that $6! = 720 = \dfrac{{240}}{3}$ , so all the terms after $6!$ including $6!$ will be divisible by $240$ (as the terms after it will contain $6!$ in their expansion). Thus the remainder is $1! + 2! + 3! + 4! + 5! = 153$