The least non-negative remainder when 548 is divided by 7 is

The correct answer is Option (1) → 1

The key is a small power of 5 that leaves remainder 1 on division by 7.

    \( \displaystyle 5^2=25\equiv 4\pmod 7 \)
    \( \displaystyle 5^3=5\times 4=20\equiv -1\pmod 7 \)

    \( \displaystyle 5^6=(5^3)^2\equiv(-1)^2=1\pmod 7 \)

Since 48 = 6 × 8,

    \( \displaystyle 5^{48}=(5^6)^8\equiv 1^8=1\pmod 7 \)

The least non-negative remainder is 1.

Shortcut: By Fermat's little theorem, \( a^{p-1}\equiv 1\pmod p \) for a prime p not dividing a. Here \( 5^6\equiv 1\pmod 7 \), and 48 is a multiple of 6, so the remainder is 1 at once.