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

Introduction / Context:
This number theory question combines the concept of least common multiple (L.C.M.) with congruences and an additional condition that the resulting number must be a multiple of 7. It tests your ability to interpret and manipulate modular arithmetic conditions in a structured way.

Given Data / Assumptions:

• We seek a number N that is a multiple of 7.
• N leaves a remainder of 4 when divided by 6, 9, 15, and 18.
• We must find the least such multiple of 7.

Concept / Approach:
If N gives remainder 4 when divided by each of 6, 9, 15, and 18, then N - 4 is exactly divisible by those numbers. Therefore, N - 4 must be a common multiple of 6, 9, 15, and 18. The smallest such value is their L.C.M. Let that L.C.M. be L. Then N can be written as N = 4 + kL. We must choose the smallest such N that is also a multiple of 7, that is, 4 + kL must be divisible by 7 for some positive integer k.

Step-by-Step Solution:
Step 1: Let N be the required number.Step 2: N leaves remainder 4 when divided by 6, 9, 15, and 18, so N - 4 is divisible by all four numbers.Step 3: Compute L = L.C.M.(6, 9, 15, 18).Step 4: Factorize: 6 = 2 * 3, 9 = 3^2, 15 = 3 * 5, 18 = 2 * 3^2.Step 5: Take highest powers: 2^1, 3^2, and 5^1.Step 6: L = 2 * 9 * 5 = 90.Step 7: So N - 4 = 90k, and N = 4 + 90k for some integer k.Step 8: We also need N to be a multiple of 7, so 4 + 90k must be divisible by 7.Step 9: Work modulo 7: 90 ≡ 6 (mod 7), so 4 + 90k ≡ 4 + 6k (mod 7).Step 10: We require 4 + 6k ≡ 0 (mod 7), or 6k ≡ -4 ≡ 3 (mod 7).Step 11: The inverse of 6 mod 7 is 6, so k ≡ 3 * 6 ≡ 18 ≡ 4 (mod 7).Step 12: The smallest positive k that satisfies this is k = 4.Step 13: Therefore N = 4 + 90 * 4 = 4 + 360 = 364.

Verification / Alternative check:
Check divisibility: 364 is a multiple of 7 (7 * 52 = 364).Check remainders: 364 ÷ 6 = 60 remainder 4, 364 ÷ 9 = 40 remainder 4, 364 ÷ 15 = 24 remainder 4, 364 ÷ 18 = 20 remainder 4. All conditions are satisfied, and because we used the smallest k compatible with the modular condition, 364 is the least such multiple.

Why Other Options Are Wrong:
Options a (74), b (94), and c (184) either do not satisfy the remainder condition for all four divisors or are not multiples of 7. A quick check reveals that at least one of the specified divisibility conditions fails for each of these numbers.

Common Pitfalls:
Many students try random trial and error with the answer choices rather than using a systematic L.C.M. and modular arithmetic approach. Others may forget that N must also be a multiple of 7 or mis-handle the modulo calculations. Carefully translating the conditions into equations and congruences helps avoid these mistakes.

Final Answer:
The least multiple of 7 that satisfies the given conditions is 364.