Question

The remainder obtained when $1! + 2! + ...... + 50!$ is divided by $30$ is
(A) $3$
(B) $13$
(C) $11$
(D) $9$

Answer 1

In order to solve this question, we must know about the concept of factorial i.e., $n! = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot ..... \cdot 3 \cdot 2 \cdot 1$ .Here, in this question, we have to find the remainder of the given factorial sum when divided by $30$ .For this, we know that the value of factorial after $5!$ will leave the remainder zero when it is divided by $30$ .So for calculating the remainder, we will calculate the remainder before that number.

Complete step by step answer:
So, first of all we will calculate the factorial till $5!$ .
So, $1! = 1$
$2! = 2 \times 1 = 2$
$3! = 3 \times 2 \times 1 = 6$
$4! = 4 \times 3 \times 2 \times 1 = 24$
and if we see $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ which is divisible by $30$
Now after $5!$ we can write each factorial term as $5!{\text{ }} \times {\text{ }}y$ where $y$ is a positive number.
i.e., $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 6 \times 5!$
$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 7 \times 6 \times 5!$ and so on.
It means the value of factorial after $5!$ will leave the remainder zero when it is divided by $30$
So, we will calculate the value remainder before $5!$
For this, add all the factorial values from $1$ to $4$
$\Rightarrow 1! + 2! + 3! + 4!$
Now on dividing by $30$ we get
$\dfrac{{1! + 2! + 3! + 4!}}{{30}}{\text{ }} - - - \left( 1 \right)$
On substituting the values, we will get the equation $\left( 1 \right)$ as
$\dfrac{{1 + 2 + 6 + 24}}{{30}}{\text{ }} - - - \left( 2 \right)$
Now adding the numerator of the equation $\left( 2 \right)$ we get,
$\dfrac{{33}}{{30}}$
And now on dividing it by $30$ we get the remainder as $3$
$\left[ {\because 33 = 30 \times 1 + 3} \right]$
Therefore, the remainder for the factorial $1! + 2! + ...... + 50!$ when divided by $30$ is $3$

Hence, the correct answer is option (A).

Note:
To solve this question, we don’t need to add all the factorial terms and divide it by $30$ to obtain the answer. Because if we see $30$ can be factored as a product of $2,3$ and $5$ . And, the least value of factorial which contains $2,3$ and $5$ is $5!$ .So $5!$ and any factorial greater than $5$ always contains $2,3$ and $5$ and hence the remainder is going to be zero for any $x!$ where $x$ is either $5$ or greater than $5$ . So, we just calculate the remainder before $5!$ and get the required result.