Find the least multiple of 7 that leaves a remainder of 4 when divided by 6, 9, 15, and 18.

Concept / Approach

• If N leaves remainder 4 upon division by 6, 9, 15, and 18, then N ≡ 4 (mod LCM(6,9,15,18)).
• Simultaneously, N must be a multiple of 7.

Step-by-step calculationLCM(6,9,15,18) = 90N ≡ 4 (mod 90) and N ≡ 0 (mod 7)Let N = 4 + 90k. Require 4 + 90k ≡ 0 (mod 7)90 ≡ 6 (mod 7) ⇒ 4 + 6k ≡ 0 ⇒ 6k ≡ 3 (mod 7)Inverse of 6 (mod 7) is 6 ⇒ k ≡ 3×6 ≡ 18 ≡ 4 (mod 7)Smallest k = 4 ⇒ N = 4 + 90×4 = 364

Verification364 ÷ 7 = 52 (multiple), and 364 leaves remainder 4 upon division by 6, 9, 15, and 18.

Final Answer364