Question

What is the remainder when $1!+2!+3!+......+2014!$ is divided by 21?
(a) 7
(b) 5
(c) 12
(d) 17

Answer 1

Hint: We start solving the problem by recalling the definition of factorial (!) of a given number. We then factorize the given divisor 21. We then check the terms which contain the product of the factors of the given divisor 21. We then find the sum of the terms which doesn’t have the product of factors of 21. We divide this sum to get the remainder, which will be the required answer for the problem.

Complete step-by-step answer:
According to the problem, we need to find the remainder obtained when $1!+2!+3!+......+2014!$ is divided by 21.
Let us recall the definition factorial (!). We know that $n!$ is defined as the product of all numbers that were less than or equal to n.
So, we have $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
Let us factorize the number 21. We can see that 21 can be written as $21=7\times 3$.
We can see that the terms after $6!$ contains multiplication of the terms 7 and 3.
So, we get the remainder from dividing the sum $1!+2!+3!+4!+5!+6!$.
Let us first find the sum and assume it as S.
So, we get $S=1+2+6+24+120+720$.
$\Rightarrow S=873$.
Now, let us divide the sum ‘S’ obtained with 21 to get the remainder.
The division process is as shown below.
$\begin{align}
  & \left. 21 \right)873\left( 41 \right. \\
 & \underline{\text{ }84\text{ }} \\
 & \text{ }33 \\
 & \underline{\text{ 21 }} \\
 & \text{ }12 \\
\end{align}$.
From the division process, we can see that the remainder is 12.
∴ The remainder is 12 when $1!+2!+3!+......+2014!$ is divided by 21.

So, the correct answer is “Option (c)”.

Note: We need not expand every factorial given as we need the remainder of the given sum. We just need to find the terms in the sum that are exactly divisible by 21 which will be neglected as we know that the sum of those terms will not increase the remainder of the sum. Whenever we get this type of problem, we first need to find the terms that contain the product of the factors of the given divisor. Similarly, we can expect problems to find the number present in the unit place for the given sum.