Find the smallest number which, when diminished by 7, is divisible by 12, 16, 18, 21, and 28.

Given data

• We seek the least N such that N − 7 is divisible by 12, 16, 18, 21, and 28 simultaneously.

Concept / Approach

• Let L = L.C.M.(12, 16, 18, 21, 28). Then N − 7 must be a multiple of L ⇒ N = Lk + 7. Minimal N occurs at k = 1.

Compute L.C.M.

Prime forms: 12 = 2^2 × 3; 16 = 2^4; 18 = 2 × 3^2; 21 = 3 × 7; 28 = 2^2 × 7.Take highest powers: 2^4, 3^2, 7^1 ⇒ L = 16 × 9 × 7 = 1008.

Smallest N

N = 1008 × 1 + 7 = 1015.

Verification

N − 7 = 1008, clearly divisible by 12, 16, 18, 21, and 28.

Common pitfalls

• Answering 1008 (forgetting to add 7 back).
• Missing 3^2 in L.C.M. due to 18.

Final Answer

Smallest such number = 1015.