On dividing a number by 56, we get 29 as remainder. On dividing the same…

Concept:

If a number N leaves remainder r when divided by d₁, and a smaller divisor d₂ exactly divides d₁ (d₁ is a multiple of d₂), then N divided by d₂ leaves the same remainder as r divided by d₂ — because the multiple-of-d₁ part of N is automatically a multiple of d₂ too, contributing nothing to the remainder.

Application:

1. Let the number be N. Since N leaves remainder 29 on dividing by 56, write N = 56Q + 29 for some whole number Q.

2. 56 is an exact multiple of 8 (56 = 8 × 7), so 56Q is also an exact multiple of 8 — it leaves remainder 0 when divided by 8.

3. So the remainder of N on dividing by 8 is the same as the remainder of 29 on dividing by 8.

4. Divide 29 by 8: 8 × 3 = 24, and 29 − 24 = 5, so 29 divided by 8 gives quotient 3 and remainder 5.

Cross-check:

Take Q = 1, so N = 56 + 29 = 85. Check: 85 divided by 56 gives quotient 1 and remainder 29 (correct). Now divide 85 by 8: 8 × 10 = 80, and 85 − 80 = 5, confirming the remainder is 5.