If 7 + y = 4 (mod 8), find the least value of y, 10 $\leq y \leq 30$

Analyze the question: We need y such that 7 + y ≡ 4 (mod 8), with y between 10 and 30. We want the smallest such y.

Concept: a ≡ b (mod n) means a and b leave the same remainder when divided by n, i.e. (a − b) is a multiple of n.

Solution:

7 + y ≡ 4 (mod 8) y ≡ 4 − 7 (mod 8) y ≡ −3 (mod 8) y ≡ 5 (mod 8) (since −3 + 8 = 5)

So y = 5, 13, 21, 29, ...

The least value in the range 10 ≤ y ≤ 30 is 13.

Check: 7 + 13 = 20, and 20 ÷ 8 = 2 remainder 4. ✓

Check the other options:

• 11: 7 + 11 = 18, remainder 2 ✗
• 19: 7 + 19 = 26, remainder 2 ✗
• 21: 7 + 21 = 28, remainder 4 ✓ but it is larger than 13

So the least value is 13, which is option B.

Tip: In modular arithmetic, first reduce to the smallest positive residue (here y ≡ 5), then keep adding the modulus (8) until you enter the allowed range. Always verify by plugging back in.