The least non-negative remainder when \( 3^{2026} \) is divided by 13 is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
A small power of 3 leaves remainder 1 on division by 13:
\( \displaystyle 3^3=27=2\times 13+1 \)
\( \displaystyle 3^3\equiv 1\pmod{13} \)
Dividing the exponent by 3, 2026 = 3 × 675 + 1, so
\( \displaystyle 3^{2026}=\left(3^3\right)^{675}\cdot 3 \)
\( \displaystyle \equiv 1^{675}\cdot 3\equiv 3\pmod{13} \)
The remainder is 3.
Why the other options are incorrect
Option (1) 9 — Incorrect. 9 is the remainder of \( 3^2 \), which would need 2026 to leave remainder 2 on division by 3; it leaves 1.
Option (2) 7 — Incorrect. 7 is not a remainder of any power of 3 on division by 13 (the remainders repeat 3, 9, 1).
Option (3) 1 — Incorrect. 1 would need 2026 to be a multiple of 3; 2026 = 3 × 675 + 1.