Problem No. 55

Number Theory Level 3

Find the last two digits of 0 ! + 5 ! + 10 ! + 15 ! + … + 100 ! . 0! + 5! + 10! + 15! + \ldots + 100!.


The answer is 21.

Benedict Dimacutac by Benedict Dimacutac , 21, Philippines

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1 solution

Tay Yong Qiang
Aug 19, 2015

10 ! = 1 × 2 × . . . × 5 × . . . × 10 = 100 M 10!=1×2×...×5×...×10=100M for some M

Factorials greater than 10 ! 10! will thus have at least 2 trailling zeroes.

The last 2 digits of the sum is thus the last 2 digits of 0 ! + 5 ! 0!+5! , which is 21 \boxed{21} .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

I think you have done a mistake when writing the solution. It'd be last 2 digits of 0 ! + 5 ! 0!+5! , rather than 5 ! + 10 ! 5!+10! .

MD Omur Faruque - 5 years, 9 months ago

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Oh yes, i've edited it.

Tay Yong Qiang - 5 years, 9 months ago

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