When 31989 is divided by 7, the remainder is:

Concept

For powers evaluated modulo a number, residues repeat in a cycle. If a power returns residue 1 after k steps, the exponent may be reduced modulo k.

For modulus 7, the powers of 3 have period 6 because 36 ≡ 1 (mod 7).

Application

1. Write the exponent as 1989 = 6 × 331 + 3, so its remainder on division by 6 is 3.

2. Therefore, 31989 ≡ 33 (mod 7).

3. Compute 33 = 27 = 7 × 3 + 6, giving residue 6.

Cross-check

The residue cycle is 3, 2, 6, 4, 5, 1 and then repeats. An exponent congruent to 3 modulo 6 therefore lands on residue 6.

Hence, the required remainder is 6.