The least number which when divided by 5, 6, 7 and 8 that leaves a remainder…

Concept: When a number leaves the same remainder r on division by several numbers, it must be of the form (LCM of those numbers) × k + r, for some whole number k. To pin down k, apply any additional condition (such as divisibility by another number) to this general expression and solve for the smallest valid k.

1. Find the LCM of 5, 6, 7 and 8. Since 8 = 23, 6 = 2 × 3, and 7 is prime, LCM(5, 6, 7, 8) = 23 × 3 × 5 × 7 = 840.

2. So the required number has the form 840k + 4, which automatically leaves remainder 4 on division by each of 5, 6, 7 and 8.

3. Impose the second condition — divisibility by 11. Since 840 = 76 × 11 + 4, 840 leaves remainder 4 on division by 11, so 840k + 4 is congruent to 4k + 4, i.e. 4(k + 1), modulo 11.

4. For 840k + 4 to be divisible by 11, 4(k + 1) must be divisible by 11. As the greatest common divisor of 4 and 11 is 1, this needs k + 1 to be divisible by 11, i.e. k ≡ 10 (mod 11). The smallest such k is 10.

5. Substituting k = 10: 840 × 10 + 4 = 8404.

Cross-check: 8404 − 4 = 8400 = 840 × 10, which is exactly divisible by 5, 6, 7 and 8, confirming the remainder-4 condition; and 8404 ÷ 11 = 764 exactly, confirming divisibility by 11.