Mod 6

Number Theory Level 1

Four positive integers a a , b b , c c , and d d are such that

{ ( a + b + c + d ) m o d 6 = 1 ( a + b + c ) m o d 6 = 0 ( a + b ) m o d 6 = 0 a m o d 6 = 3 \large \begin{cases} \begin{aligned} (a+b+c+d) \bmod 6 & = 1 \\ (a+b+c) \bmod 6 & = 0 \\ (a+b) \bmod 6 & = 0 \\ a \bmod 6 & = 3 \end{aligned} \end{cases}

What is ( 2 ( a − d ) + c 2 − 3 b ) m o d 6 \big(2(a-d)+c^2-3b\big) \bmod 6 ?

Try this


The answer is 1.

ابراهيم فقرا by ابراهيم فقرا , 23, Israel

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2 solutions

Jordan Cahn
Dec 21, 2018

Using the given clues, from the bottom up, we see that

  • a ≡ 3 m o d 6 a\equiv 3\bmod 6
  • b ≡ 3 m o d 6 b\equiv 3\bmod 6
  • c ≡ 0 m o d 6 c\equiv 0\bmod 6
  • d ≡ 1 m o d 6 d\equiv 1\bmod 6

Thus 2 ( a − d ) + c 2 − 3 b ≡ 2 ( 3 − 1 ) + 0 2 − 3 ( 3 ) ≡ 4 − 9 ≡ 1 m o d 6 2(a-d) + c^2 -3b \equiv 2(3-1) + 0^2 -3(3) \equiv 4 - 9 \equiv 1 \mod 6

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Chew-Seong Cheong
Dec 21, 2018

a ≡ 3 (mod 6) Given a + b ≡ 3 + b ≡ 0 (mod 6) ⟹ b ≡ 3 (mod 6) a + b + c ≡ 0 + c ≡ 0 (mod 6) ⟹ c ≡ 0 (mod 6) a + b + c + d ≡ 0 + d ≡ 1 (mod 6) ⟹ d ≡ 1 (mod 6) \begin{aligned} a & \equiv 3 \text{ (mod 6)} & \small \color{#3D99F6} \text{Given} \\ a+b & \equiv 3 + b \equiv 0 \text{ (mod 6)} & \small \color{#3D99F6} \implies b \equiv 3 \text{ (mod 6)} \\ a+b+c & \equiv 0 + c \equiv 0 \text{ (mod 6)} & \small \color{#3D99F6} \implies c \equiv 0 \text{ (mod 6)} \\ a+b+c + d & \equiv 0 + d \equiv 1 \text{ (mod 6)} & \small \color{#3D99F6} \implies d \equiv 1 \text{ (mod 6)} \end{aligned}

Therefore, 2 ( a − d ) + c 2 − 3 ≡ 2 ( 3 − 1 ) + 0 2 − 3 ≡ 1 (mod 6) 2(a-d) + c^2 - 3 \equiv 2(3-1)+0^2 - 3 \equiv \boxed 1 \text{ (mod 6)} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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