Let x be exactly divisible by 5 (remainder 0) and let y leave a remainder of 1 when divided by 5. When (x + y) is divided by 5, what remainder is obtained?

Introduction / Context:
Remainder or modular arithmetic problems often ask how remainders combine under addition. If one addend is a multiple of 5 and the other leaves remainder 1 modulo 5, the sum’s remainder follows directly from basic modular properties.

Given Data / Assumptions:

• x ≡ 0 (mod 5), meaning x is a multiple of 5.
• y ≡ 1 (mod 5), meaning dividing y by 5 leaves remainder 1.
• We need (x + y) mod 5.

Concept / Approach:
Use the modular identity: (a + b) mod m = [(a mod m) + (b mod m)] mod m. This rule allows us to compute the remainder of a sum from the remainders of the terms. No explicit values for x or y are needed, only their equivalence classes modulo 5.

Step-by-Step Solution:

Verification / Alternative check:
Pick examples: let x = 10 (clearly ≡ 0 mod 5) and y = 11 (≡ 1 mod 5). Then x + y = 21, and 21 divided by 5 leaves remainder 1. Any other valid pair will behave the same due to modular equivalence.

Why Other Options Are Wrong:
0, 2, 3, and 4 contradict the direct application of modular addition. They typically arise from guessing or ignoring the given remainder for y.

Common Pitfalls:
Assuming that any multiple of 5 changes the remainder of the sum (it does not). Another mistake is to add quotients rather than remainders.

Final Answer:
1