Remainder problem — What is the remainder when 19^100 is divided by 20? Use the symmetry 19 ≡ −1 (mod 20).

Introduction / Context:
Powers of numbers that are “one less than” the modulus often simplify neatly. Here 19 ≡ −1 (mod 20), making high powers easy to evaluate using parity. This technique is a must-know for number system questions involving large exponents and small moduli.

Given Data / Assumptions:

• Compute 19^100 mod 20.
• Observation: 19 ≡ −1 modulo 20.
• Even powers of −1 equal +1.

Concept / Approach:
Replace 19 with −1 under modulo 20 and apply exponent rules. Because the exponent 100 is even, the result collapses to 1 immediately, without any need for cycles or Euler's theorem.

Step-by-Step Solution:
1) 19 ≡ −1 (mod 20).2) 19^100 ≡ (−1)^100 (mod 20).3) Since 100 is even, (−1)^100 = 1.4) Therefore, the remainder is 1.

Verification / Alternative check:
Euler’s theorem with φ(20) = 8 also works but is unnecessary. The parity shortcut is the simplest and least error-prone approach here.

Why Other Options Are Wrong:
19 would correspond to an odd power; 3 or 29 do not arise from the parity shortcut; 0 is impossible because 20 does not divide 19.

Common Pitfalls:
Mixing up odd/even exponents; misapplying Euler’s theorem; forgetting that the remainder must be in 0–19 inclusive.

Final Answer:
1