Find the remainder: (17 13 − 21) ÷ 18

Correct — A. Work modulo 18, where the base is one less than the divisor.

17 ≡ −1 (mod 18), since 17 = 18 − 1
17¹³ ≡ (−1)¹³ = −1 (mod 18), the power being odd
17¹³ − 21 ≡ −1 − 21 = −22 (mod 18)

A remainder has to land in 0…17, so add 18 twice: −22 + 36 = 14 → option (a).

Why the others are wrong

• (b)
  21 — 21 is larger than the divisor, so it cannot be a remainder on division by 18 at all — the residues run 0 to 17. It is the constant lifted straight out of the bracket.
• (c)
  15 — 15 would need the bracket to leave −3 modulo 18. It leaves −22, and −22 + 36 = 14, so no step of the working ever produces a 3.
• (d)
  17 — 17 is the remainder of 17¹³ by itself, which is only the halfway point. The − 21 still has to be applied, and it brings the answer down to 14.

When the base is one less than the divisor, the power collapses to a sign.

17 = 18 − 1, so 17 ≡ −1 (mod 18) and 17ⁿ ≡ (−1)ⁿ. An odd power leaves −1 and an even power leaves +1, so with the exponent 13 nothing has to be raised at all.

The second habit is finishing inside the legal range. A negative residue is still a residue: keep adding the divisor until the value sits between 0 and 17.

Key facts

• 17 ≡ −1 (mod 18), so 17ⁿ leaves 17 for odd n and 1 for even n.
• A remainder on division by 18 must lie between 0 and 17, which rules out any option of 18 or more.
• −22 ≡ 14 (mod 18), because −22 + 36 = 14.

Common traps

• Reading (−1)¹³ as +1, which finishes at 16 instead of 14
• Reporting the negative residue −4 instead of adding the divisor to reach 14
• Answering 17, the remainder of the power alone, before the 21 is subtracted

No directly related past PYQ was found.