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Number Theory Level 3

Find the remainder when 1 ! + 2 ! + 3 ! + ⋯ + 2017 ! 1!+2!+3!+\cdots+2017! is divided by 7.

Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

4 3 1 Cannot be determined 6 5 0 2

I Gede Arya Raditya Parameswara by I Gede Arya Raditya Para… , 17

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2 solutions

Chew-Seong Cheong
Dec 25, 2016

Since n ! n! is divisible by 7 for n ≥ 7 n \ge 7 . The required reminder is given by:

1 ! + 2 ! + 3 ! + 4 ! + 5 ! + 6 ! ≡ 1 + 2 + 6 + 24 + 120 + 720 (mod 7) ≡ 1 + 2 + 6 + 3 + 1 + 6 (mod 7) ≡ 19 (mod 7) ≡ 5 (mod 7) \begin{aligned} 1!+2!+3!+4!+5!+6! & \equiv 1+2+6+24+120+720 \text{ (mod 7)} \\ & \equiv 1+2+6+3+1+6 \text{ (mod 7)} \\ & \equiv 19 \text{ (mod 7)} \\ & \equiv \boxed{5} \text{ (mod 7)} \end{aligned}

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7! Mod(7) is 0 So 7!+8!+9!+...+2017! Mod(7) is 0

1!+2!+...+6! Mod(7) is 5 So the answer is 5

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