Question: The tens digit of 1! + 2! + 3! + ………. + 49! is equal to? 1). 1 2). 2 3). 3 4). 4...

Question

The tens digit of 1! + 2! + 3! + ………. + 49! is equal to?
1). 1
2). 2
3). 3
4). 4

Answer 1

Solution

We have to find the values of 1! , 2! Till 9! And then, calculate the sum of unit place and tens place of these values. Whichever digit is at tens place is the final value because from 10! Onwards the value of tens place and unit place digits will be 0.

Complete step-by-step solution:
We will find the values of 1! To 9!
$1! = 1$
$2! = 2$
$3! = 6$
$4! = 24$
$5! = 120$
$6! = 720$
$7! = 5040$
$8! = 40320$
$9! = 362880$
Now, we will find the sum of unit places and tens places of all these.
$1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 = 213$
So, the sum is 213, which means the digit on tens place is 1.
The value of 10! Is 3628800 and in this there is 0 in both tens and unit places.
If the value is 10! Has 0 in both tens and unit places then the values which will come after it will obviously have 0 in tens and unit place. So, which means the value we calculated earlier of unit and tens place will remain the same.
So, this tells us that the digit on tens place will be 1 only.
Option (1) is the correct answer.

Note: Factorial is the product of all positive integers less than or equal to a given positive integer and denoted by that integer and an exclamation point. Thus, factorial seven is written as 7! Which means 1234567. Factorial zero is defined as equal to 1.