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# If $K=1!+2\times 2!+3\times 3!+....+20\times 20!$ then K + 5 when divided by 21! Leaves a remainder?(a) 0(b) 1(c) 5(d) 4

Last updated date: 22nd Jul 2024
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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.

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.