Find the smallest number to add to 2497 so the sum is divisible by 5, 6, 4, and 3.

Given data

• We need k ≥ 0 such that 2497 + k is divisible by each of 5, 6, 4, 3.

Concept / Approach

• Let L = L.C.M.(5, 6, 4, 3). Then 2497 + k ≡ 0 (mod L).

Step-by-step calculation

Prime forms: 5 = 5; 6 = 2 × 3; 4 = 2^2; 3 = 3.L = 2^2 × 3 × 5 = 60.2497 mod 60: 60 × 41 = 2460 ⇒ remainder 37.Need k ≡ −37 ≡ 23 (mod 60).Minimum such k = 23.

Verification

(2497 + 23) = 2520, which is divisible by 60 (and by 5, 6, 4, 3 individually).

Common pitfalls

• Checking divisibility by each number separately without using L.C.M., increasing error risk.

Final Answer

Required minimal addition = 23.