# The remainder obtained when $1! + 2! + ......49!$ is divided by $20$ is  $\left( a \right){\text{ 13}}$  $\left( b \right){\text{ 33}}$  $\left( c \right){\text{ 12}}$  $\left( d \right){\text{ 11}}$

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Hint: Here, in this question, we have to find the remainder of the given factorial sum. For this, we know that the value of factorials after $5!$ will leave the remainder zero when it is divided by $20$ . So for calculating the remainder, we will calculate the remainder before that number.

So first of all we will calculate the factorial till $5!$ . So for this, the factorials will be
$1! = 1$ , $2! = 2 \times 1 = 2$ , $3! = 3 \times 2 \times 1 = 6$ , $4! = 4 \times 3 \times 2 \times 1 = 24$
And the factorial after $5!$ will have the remainder zero. So now adding all the factorial values from $1$ to $5$ , we get
$\Rightarrow \dfrac{{1! + 2! + 3! + 4!}}{{20}}$
And on substituting the values, we had obtained above, we will get the equation as
$\Rightarrow \dfrac{{1 + 2 + 6 + 24}}{{20}}$
Now adding the numerator of the above fraction, we get the fraction as
$\Rightarrow \dfrac{{33}}{{20}}$
And now on dividing it, we have the remainder as $13$ .
Therefore, the remainder for the factorial $1! + 2! + ......49!$ when divided by $20$ is $13$ .
Hence, the option $\left( a \right)$ is correct.
So, the correct answer is “Option a”.

Note: $20$ can be factored as a product of $5$ and $4$ . So the least value of factorial which contains both $5$ and $4$ is $5!$ . So $5!$ and any factorial greater than $5$ always contains both $5$ and $4$ , so the remainder is going to be zero for any $x!$ . Where $x$ is either $5$ or greater than $5$ .
For solving this type of question we should know how to calculate the factorial of any number. And also if we memorize the properties then we don’t have to solve and add that long factorial value. By using $n! = n \times \left( {n - 1} \right)!$ , here $n$ is the number whose factorial is to be calculated and in this way, we will get the factorial of any number.