Find the remainder when the number (218 + 5) is divided by 9.

Concept: To find the remainder of a large power expression modulo a divisor, first reduce the base to a small residue — ideally ±1 — modulo that divisor, then raise ONLY the residue to the power. A base congruent to −1 (mod n) raised to an EVEN power is congruent to 1 (mod n), and to an ODD power is congruent to −1 (mod n).

Application:

1. Write 23 = 8. Since 9 − 8 = 1, we have 8 ≡ −1 (mod 9).

2. Since 18 = 3 × 6, rewrite 218 as (23)6 = 86.

3. Substitute the congruence 8 ≡ −1 (mod 9): 86 ≡ (−1)6 (mod 9).

4. Because the exponent 6 is even, (−1)6 reduces to 1, so 218 ≡ 1 (mod 9).

5. Add the 5 from the original expression: 218 + 5 ≡ 1 + 5 = 6 (mod 9).

6. Since 6 is already smaller than the divisor 9, 6 itself is the final remainder.

Cross-check: The multiplicative order of 2 modulo 9 is 6, since 26 = 64 = 63 + 1 ≡ 1 (mod 9). As 18 is itself a multiple of 6, 218 = (26)3 ≡ 13 = 1 (mod 9), the same residue found above — confirming the remainder.

Result: The remainder when (218 + 5) is divided by 9 is 6.