Find the least number greater than 35 which, when divided by 12, 27 and 35,…

Concept: If a number N leaves the same remainder r when divided by several divisors, then N − r is exactly divisible by every one of those divisors — that is, N − r is a common multiple of all the divisors. The least N greater than the divisors is obtained by taking N − r equal to the LCM of the divisors, so N = LCM(divisors) + r.

Application:
Here the divisors are 12, 27 and 35, and the remainder is r = 6. The stem requires N to be greater than 35, so N − 6 must be a positive common multiple of 12, 27 and 35 — the least such N uses the LCM.

1. Set up the condition: if N is the required number, N − 6 is divisible by 12, 27 and 35.

2. Prime-factorize each divisor: 12 = 22 × 3, 27 = 33, 35 = 5 × 7.

3. Take the highest power of each prime to form the LCM: LCM = 22 × 33 × 5 × 7 = 4 × 27 × 35 = 3780.

4. Add the remainder back: N = 3780 + 6 = 3786.

Cross-check: Verify 3786 directly against each divisor — 3786 ÷ 12 = 315 remainder 6; 3786 ÷ 27 = 140 remainder 6; 3786 ÷ 35 = 108 remainder 6. Every division leaves exactly remainder 6, and 3786 is greater than 35, confirming it is the least number satisfying all the stated conditions.