Problem 4: Large factorials

Level 2

− 132 ! + 131 ! − 128 ! 128 ! ≡ x (mod 137) \large \frac{-132!+131!-128!}{128!} \equiv x \text{ (mod 137)}

Find x x , where 0 ≤ x ≤ 136 0 \le x \le 136 .


The answer is 38.

Ashwin R by Ashwin R , 16, USA

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1 solution

Chew-Seong Cheong
Apr 25, 2018

− 128 ! + 131 ! − 128 ! 128 ! ≡ 128 ! ( − 129 ⋅ 130 ⋅ 131 ⋅ 132 + 129 ⋅ 130 ⋅ 131 − 1 ) 128 ! (mod 137) ≡ 129 ⋅ 130 ⋅ 131 ( − 132 + 1 ) − 1 (mod 137) ≡ − 129 ⋅ 130 ⋅ 13 1 2 − 1 (mod 137) ≡ − ( − 8 ) ( − 7 ) ( − 6 ) 2 − 1 (mod 137) ≡ − 56 ⋅ 36 − 1 (mod 137) ≡ − 2017 (mod 137) ≡ − 99 (mod 137) ≡ 38 (mod 137) \begin{aligned} \frac {-128!+131!-128!}{128!} & \equiv \frac {128!(-129\cdot 130 \cdot 131 \cdot 132 + 129 \cdot 130 \cdot 131 - 1)}{128!} \text{ (mod 137)} \\ & \equiv 129\cdot 130 \cdot 131 (-132+1) - 1 \text{ (mod 137)} \\ & \equiv - 129\cdot 130 \cdot 131^2 - 1 \text{ (mod 137)} \\ & \equiv -(-8)(-7)(-6)^2 - 1 \text{ (mod 137)} \\ & \equiv -56\cdot 36 - 1 \text{ (mod 137)} \\ & \equiv - 2017 \text{ (mod 137)} \\ & \equiv - 99 \text{ (mod 137)} \\ & \equiv \boxed{38} \text{ (mod 137)} \end{aligned}

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