If a positive integer n leaves a remainder of 2 when divided by 5, which of the following statements must always be true?

Introduction / Context:
This question tests understanding of modular arithmetic and how remainders behave when we add constants to a number. We are told how n behaves when divided by 5 and asked which statement must hold for all such integers n. The idea is to write n in a standard modular form and examine how expressions like n + 1 or n + 3 behave.

Given Data / Assumptions:
n is a positive integer.When n is divided by 5, the remainder is 2.Symbolically, this means n ≡ 2 (mod 5).We must identify which statement is always true for every such n.

Concept / Approach:
If n leaves remainder 2 when divided by 5, we can write n in the form n = 5k + 2 for some integer k. We then substitute this expression into each option. For each statement, we check whether it holds for all integers k or fails for at least one choice of k. A statement is a must-true property only if it holds for every possible k.

Step-by-Step Solution:
Step 1: Write n as n = 5k + 2, where k is an integer.Step 2: Examine option D, which states that n + 3 is divisible by 5.Step 3: Compute n + 3 = (5k + 2) + 3 = 5k + 5.Step 4: Factor this as 5(k + 1). Since k + 1 is an integer, n + 3 is always a multiple of 5.Step 5: Therefore, n + 3 is divisible by 5 for all integers k.

Verification / Alternative check:
Test a few specific values. For k = 1, n = 7. Then n + 3 = 10, which is divisible by 5. For k = 3, n = 17, and n + 3 = 20, again divisible by 5. For k = 5, n = 27, and n + 3 = 30, still divisible by 5. These examples support the algebraic reasoning that n + 3 is always divisible by 5.

Why Other Options Are Wrong:
The statement that n is odd is not always true. For k = 2, n = 12, which is even. The claim that n + 1 cannot be a prime is false because for k = 2, n + 1 = 13, which is a prime number. The statement about (n + 2) divided by 7 having remainder 2 is not guaranteed; the remainder depends on k and can vary, so it is not a must-true condition.

Common Pitfalls:
A common mistake is to test only one or two values and assume a pattern without algebraic proof. Another error is to confuse what is sometimes true with what must always be true. Writing n as 5k + 2 and analyzing each option symbolically ensures the conclusion is valid for all possible integers n.

Final Answer:
The statement that must always hold is that n + 3 is divisible by 5.