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

Introduction / Context:
This is a simultaneous congruences problem. Because the required remainder is the same (4) for all divisors, we can combine the conditions using the least common multiple (LCM). We also require the result to be a multiple of 7.

Given Data / Assumptions:

• N ≡ 4 (mod 6), N ≡ 4 (mod 9), N ≡ 4 (mod 15), N ≡ 4 (mod 18)
• N is a multiple of 7

Concept / Approach:
First, condense the congruences to N ≡ 4 (mod LCM(6, 9, 15, 18)). Then among numbers of the form 4 + k*LCM, find the least one divisible by 7.

Step-by-Step Solution:
LCM(6, 9, 15, 18) = 2 * 3^2 * 5 = 90.General form: N = 4 + 90t.Require 7 | N ⇒ 4 + 90t ≡ 0 (mod 7). Since 90 ≡ 6 (mod 7), we need 4 + 6t ≡ 0 ⇒ 6t ≡ 3 ⇒ t ≡ 4 (mod 7).Smallest t = 4 ⇒ N = 4 + 90*4 = 364.

Verification / Alternative check:
364 mod 6, 9, 15, 18 equals 4; and 364/7 = 52 is integral. Conditions satisfied.

Why Other Options Are Wrong:
74, 94, 184, 454 do not satisfy both constraints simultaneously (either not ≡ 4 modulo the LCM or not a multiple of 7).

Common Pitfalls:
Forgetting to combine the congruences using the LCM; testing only one modulus; or missing the multiple-of-7 requirement.

Final Answer:
364