A number when divided by 5, 9, 13 leaves remainder 2, 6 and 10 respectively.…

Concept

When the divisors and remainders have a CONSTANT difference — that is, (divisor − remainder) is the same for every condition — the number is just "short of a common multiple". If N leaves remainder r when divided by d, and (d − r) = k is the same for all divisors, then N + k is exactly divisible by every divisor. Hence N + k = LCM(divisors) × t, and the least value comes from t = 1.

Application

Check the gaps (divisor − remainder) for each condition:

1. 5 − 2 = 3

2. 9 − 6 = 3

3. 13 − 10 = 3

The gap is the same (k = 3) in every case. So N + 3 is divisible by 5, 9 and 13 together.

1. LCM(5, 9, 13): these are pairwise co-prime, so LCM = 5 × 9 × 13 = 585.

2. N + 3 = 585 × t. Least value at t = 1 gives N + 3 = 585.

3. N = 585 − 3 = 582.

Cross-check

Verify 582 against all three conditions directly:

• 582 ÷ 5 = 116 remainder 2 ✓

• 582 ÷ 9 = 64 remainder 6 ✓

• 582 ÷ 13 = 44 remainder 10 ✓

All three remainders match, so the least such number is 582. (The next such number would be 585 × 2 − 3 = 1167.)