Remainder problem — Find the remainder when 17^200 is divided by 18. Use modular patterns or parity arguments for efficiency.

Introduction / Context:
Large exponents are best handled via modular arithmetic patterns. Here, 17 is close to 18, which offers a quick simplification. Recognizing small congruences prevents unnecessary exponentiation and speeds up remainder calculations dramatically.

Given Data / Assumptions:

• Compute 17^200 mod 18.
• Use congruences and exponent rules.
• We only need the remainder (0–17).

Concept / Approach:
Observe that 17 ≡ −1 (mod 18). Then 17^200 ≡ (−1)^200 (mod 18). Since any even power of −1 equals 1, the remainder is immediate. No need for Euler’s theorem here, though it would also work.

Step-by-Step Solution:
1) Reduce base: 17 ≡ −1 (mod 18).2) Raise to the 200th power: (−1)^200 = 1.3) Therefore, 17^200 ≡ 1 (mod 18).4) Remainder upon division by 18 is 1.

Verification / Alternative check:
Euler’s theorem: φ(18) = 6, so 17^6 ≡ 1 (mod 18). Then 17^200 = (17^6)^{33} * 17^2 ≡ 1^{33} * 17^2 ≡ (−1)^2 = 1, confirming the same result.

Why Other Options Are Wrong:
3, 4, 5, and 7 do not match the clean parity conclusion (even power of −1). Any non-1 remainder would contradict the simple congruence.

Common Pitfalls:
Expanding 17^200 directly; forgetting that even powers of −1 yield +1; misapplying Euler's theorem without reducing correctly.

Final Answer:
1