Easier than it looks

Number Theory Level 3

Find the last 2 digits of ∑ r = 1 100 r ! \sum\limits_{r=1}^{100} r!


The answer is 13.

Josh Banister by Josh Banister , 24, United Kingdom

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1 solution

Paola Ramírez
Jul 14, 2015

The last digit is defined by the last digit of the sum 1 ! + 2 ! + 3 ! + 4 ! = 3 1!+2!+3!+4!=3 , and the second-last by the second-lasts sum of 4 ! + 5 ! + 6 ! + 7 ! + 8 ! + 9 ! + carry = 1 4!+5!+6!+7!+8!+9! + \text{carry}=1

∴ \therefore the last 2 2 digits are: 13 \boxed{13}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Very nice but how do you know the number n ! n! where n ≥ 10 n\geq 10 won't affect the value?

Josh Banister - 5 years, 11 months ago

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Since n ≥ 10 n\geq 10 the second-last number of each n ! n! will be zero because they have two 10 10 factors. The same apply if we want to find the 3 , 4 , 5... 3,4,5... digits, we'll have to look when the n ! n! has 3 , 4 , 5... 3,4,5... factors 10 10

Paola Ramírez - 5 years, 11 months ago

How did you calculate the last digit Can you explain

A Former Brilliant Member - 4 years, 7 months ago

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