Easy factorials and modular arithmetic

What is 2019 ! + 2018 ! + 2017 ! + . . . + 3 ! + 2 ! + 1 ! ( m o d 100 ) 2019! + 2018! +2017! +...+ 3! +2! +1! (\bmod100) ?


The answer is 13.

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2 solutions

Chew-Seong Cheong
Feb 28, 2019

N 1 ! + 2 ! + 3 ! + + 2019 ! (mod 100) Since n ! m o d 100 = 0 for n 10 1 ! + 2 ! + 3 ! + 4 ! + 5 ! + 6 ! + 7 ! + 8 ! + 9 ! (mod 100) 1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 (mod 100) 1 + 2 + 6 + 4 (mod 100) 13 (mod 100) \begin{aligned} N & \equiv 1! + 2! + 3! + \cdots + 2019! \text{ (mod 100)} & \small \color{#3D99F6} \text{Since }n! \bmod 100 = 0 \text{ for }n \ge 10 \\ & \equiv 1! + 2! + 3! + 4!+5!+6!+7!+8! + 9! \text{ (mod 100)} \\ & \equiv 1 + 2 + 6 + 24+ 20 +20+40+20 + 80 \text{ (mod 100)} \\ & \equiv 1 + 2 + 6 + 4 \text{ (mod 100)} \\ & \equiv \boxed{13} \text{ (mod 100)} \end{aligned}

Nice ! Thank you for posting a solution :)

Abd FmL<3 - 2 years, 3 months ago

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You are welcome.

Chew-Seong Cheong - 2 years, 3 months ago
Kyle T
Feb 28, 2019

You should try it yourself :)

Abd FmL<3 - 2 years, 3 months ago

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