Remainder with a smaller divisor from a larger modulus condition A number leaves remainder 29 when divided by 56. Find the remainder when the same number is divided by 8.

Introduction / Context: When you know a number's remainder modulo a larger divisor, you can find its remainder modulo any factor of that divisor by reducing the given remainder accordingly. This leverages congruence compatibility.

Given Data / Assumptions:

• N ≡ 29 (mod 56).
• Need N mod 8.
• 8 divides 56, so reduction is direct.

Concept / Approach: If N ≡ r (mod m) and k divides m, then N ≡ r (mod k) but r must be reduced modulo k to get the canonical remainder between 0 and k − 1. Compute r mod k directly.

Step-by-Step Solution:Given N ≡ 29 (mod 56).Compute 29 mod 8: 8*3 = 24; remainder = 29 − 24 = 5.Therefore, N ≡ 5 (mod 8).

Verification / Alternative check: Any number of the form N = 56q + 29 will be 8*(7q) + 5, so the remainder on division by 8 is always 5.

Why Other Options Are Wrong:4, 6, 7 are simply other residues modulo 8; only 5 matches 29 reduced modulo 8.

Common Pitfalls: Dividing 29 by 8 incorrectly; trying to recompute N explicitly instead of using modular reduction; forgetting to bring the remainder into the 0–7 range.

Final Answer: 5