If a number is divided by 357 the remainder is 5, what will be the remainder…

Concept: If a number N leaves remainder r on division by D (so N = Dq + r), and D itself is an exact multiple of a smaller divisor d (D = d × m), then N = d × (mq) + r. The term d × (mq) is exactly divisible by d, so N leaves the SAME remainder r when divided by d — provided r is already less than d.

Application:

1. Let the number be N. The given condition means N = 357k + 5 for some integer k.

2. Factor 357: 357 = 17 × 21, so 357 is an exact multiple of 17.

3. Rewrite N using this factor: N = 17 × (21k) + 5. The term 17 × (21k) is exactly divisible by 17.

4. Since the constant 5 is already less than 17, dividing N by 17 leaves this constant as the remainder directly.

Cross-check: Take k = 1, so N = 357 × 1 + 5 = 362. Dividing 362 by 17 gives 362 = 17 × 21 + 5, quotient 21 and remainder 5 — confirming the result independently of the algebraic argument.

Hence, the remainder when the number is divided by 17 is 5.