Answer

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**Hint:**Write the given expression of K as a summation series of the expression $r\times r!$ form r = 1 to r = 20. Now, leave the factorial term as it is and write the other term of r as r = (r + 1) – 1. Multiply r! with each term and use the formula $\left( n+1 \right)\times n!=\left( n+1 \right)!$ to simplify. Remove the summation sign and add the terms. Cancel the like terms and add both the sides 5 to find the expression of K + 5. Divide both the sides with 21! and see the term which is left as the remainder.

**Complete step by step answer:**

Here we have been provided with the expression $K=1!+2\times 2!+3\times 3!+....+20\times 20!$ and we are asked to find the remainder when the expression K + 5 is divided by 21!. First let us simplify the expression of K.

$\Rightarrow K=1\times 1!+2\times 2!+3\times 3!+....+20\times 20!$

In the summation form we can write the above expression as: -

$\Rightarrow K=\sum\limits_{r=1}^{20}{r\times r!}$

We can write r = (r + 1) – 1, so we get,

$\begin{align}

& \Rightarrow K=\sum\limits_{r=1}^{20}{\left[ \left( r+1 \right)-1 \right]\times r!} \\

& \Rightarrow K=\sum\limits_{r=1}^{20}{\left[ \left( r+1 \right)\times r!-1\times r! \right]} \\

& \Rightarrow K=\sum\limits_{r=1}^{20}{\left[ \left( r+1 \right)\times r!-r! \right]} \\

\end{align}$

Using the formula of factorial given as $\left( n+1 \right)\times n!=\left( n+1 \right)!$ we get,

$\Rightarrow K=\sum\limits_{r=1}^{20}{\left[ \left( r+1 \right)!-r! \right]}$

Now, removing the summation sign and substituting the values of r we get,

$\Rightarrow K=\left( 2!-1! \right)+\left( 3!-2! \right)+\left( 4!-3! \right)+.....+\left( 21!-20! \right)$

Cancelling the like term and simplifying we get,

$\begin{align}

& \Rightarrow K=21!-1! \\

& \Rightarrow K=21!-1 \\

\end{align}$

Adding 5 both the sides we get,

$\begin{align}

& \Rightarrow K+5=21!-1+5 \\

& \Rightarrow K+5=21!+4 \\

\end{align}$

Dividing both the sides with 21! We get,

$\begin{align}

& \Rightarrow \dfrac{\left( K+5 \right)}{21!}=\dfrac{\left( 21!+4 \right)}{21!} \\

& \Rightarrow \dfrac{\left( K+5 \right)}{21!}=1+\dfrac{4}{21!} \\

\end{align}$

Clearly we can see that 4 cannot be divided with 21! So 4 will be the remainder of the above expression.

**So, the correct answer is “Option d”.**

**Note:**Note that it is not possible to find the overall sum of the expression of K in numbers because it will be difficult to calculate the factorials of large numbers and that is why we need to form a pattern so that we can simplify the expression. Remember the properties of factorials and their formulas to make the simplification easier.

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