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# The remainder obtained when $1! + 2! + 3! + ........................100$ is divided by $240$ isA. $153$B. $154$C. $155$D. $156$

Last updated date: 19th Jun 2024
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Hint: As we all are very aware of factorial so in this question we will firstly try to expand our initial factorial by using one simple property that is
$n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right)............1$
So, we will use this property to find the remainder by dividing it by $240$.

Step by step solution:
So, firstly we need to find the sum of $1! + 2! + 3! + ........................100$, then by dividing it by $240$ we will get our result.
Now, we will expand the factorial
$= \dfrac{{(1) + (2 \times 1) + (3 \times 2 \times 1) + (4 \times 3 \times 2 \times 1) + (5 \times 4 \times 3 \times 2 \times 1) + (6 \times 5 \times 4 \times 3 \times 2 \times 1) + (7 \times 6!) + ...........100!}}{{240}}$
Now, as we can see that before $6$ any of the number is not divisible by $240$ but when we divide
$6!{\text{ }}from{\text{ }}240$, it get full divide without leaving any remainder that is
$= \dfrac{{153}}{{240}} + \dfrac{{720}}{{240}} + \dfrac{{7 \times 6!}}{{240}} + ....................\dfrac{{100!}}{{240}}$
So, $\dfrac{{720}}{{240}}$ is fully divisible and we get remainder 0 so after $6$ all factorial must contain the term $6$ there is no need to carry division forward, we will get the same result.
Now, if we see before $6$ upto $5!$ none of the factorial is divisible so this means that it leaves a remainder so by adding all the factorial before $6$ we get value as $153$ which is also written as remainder of this sum when divided by $240$.

Note:
While dividing any factorial by any number to find remainder we must observe the pattern that is upto which place our factorial will start giving us remainder zero. Also, we know that the factorial of $1!{\text{ }}is{\text{ }}1$, but $0$ has also a value equal to $1$.