Question

The last two digits in $X = \sum\limits_{k = 1}^{100} {k!} $ are?
$
  1)10 \\
  2)11 \\
  3)12 \\
  4)13 \\
$

Answer 1

Hint: Here the given expression is in the form of factorial. Factorial can be defined as the product of all the positive integers less than or equal to the given number. For Example if “n” is the positive integer, then “n” factorial is denoted by $n!$ it is the product of all positive integers less than or equal to n. here we will use the factorial formulas to find its summation.

Complete step-by-step answer:
Here we are asked to find the summation till the number hundred, here we will find the factorial till the number ten.
$
  1! = 1 \\
  2! = 2 \\
  3! = 6 \\
  4! = 24 \\
  5! = 120 \\
  6! = 720 \\
  7! = 5040 \\
  8! = 40320 \\
  9! = 362880 \\
  10! = 3628800 \\
$
It's worthless to find the further factorials since when the last two digits come double zero then further values would be repeated.
So, now finding the sum of the last two terms in these one to ten factorials
$ = 1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 + 00$
Simplify the above expression finding the sum of the numbers
$ = 213$
Hence, the last two digits in the summation would be equal to $13$
Hence, from the given multiple choices – fourth option is the correct answer.
So, the correct answer is “Option B”.

Note: Don’t get confused between the two terms factorization and factorial. Be good in multiples and division and remember at least twenty since the factorial of any term depends on the product of the terms. Factorization is the process of finding the numbers which are when multiplied to get the original number.