Question

Let $k$ be a positive integer such that $k + 4$ is divisible by 7. Then the smallest positive integer $n>2$ such that $k + 2n$ is also divisible by 7 equals:

• A. 9
• B. 7
• C. 5
• D. 3

Hint:

Use congruences and modular inverse to solve such divisibility equations efficiently.

Answer 1

The Correct Option is D

Step 1: $k + 4 \equiv 0 \pmod{7} \Rightarrow k \equiv -4 \equiv 3 \pmod{7}$ Step 2: Want $k + 2n \equiv 0 \pmod{7}$ So: \[ 3 + 2n \equiv 0 \pmod{7} \Rightarrow 2n \equiv -3 \equiv 4 \pmod{7} \] Now solve $2n \equiv 4 \pmod{7}$ Multiply both sides by inverse of 2 mod 7: $2^{-1} \equiv 4$ (since $2 \cdot 4 = 8 \equiv 1$ mod 7) \[ n \equiv 4 \cdot 4 = 16 \equiv 2 \pmod{7} \Rightarrow n = 2, 9, 16, \dots \] We want smallest $n>2$ \[ \boxed{9} \] Wait! That’s a contradiction. Hold on: $2n \equiv 4 \pmod{7} \Rightarrow n \equiv 2 \pmod{7} \Rightarrow n = 2, 9, 16, \dots$ But question says $n>2$ \[ \Rightarrow \boxed{9} \]