A number theory problem by Arka Das
Number Theory
Level
pending
1!+2!+...............+99!+100! what are the last two digits of there sum ?
The answer is 13.
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1 solution
Note that, for n ≥ 1 0 , n ! is divisible by 1 0 0 . Thus, we only need to calculate the last two digits of 1 ! + 2 ! + 3 ! + 4 ! + . . . + 9 ! . Taking the numbers m o d 1 0 0 , the sum is equivalent to 1 + 2 + 6 + 2 4 + 2 0 + 2 0 + 4 0 + 2 0 + 8 0 ≡ 1 3 m o d 1 0 0 .