Find the least positive integer which, when divided by each of 12, 16, and 18, leaves a remainder of 5 in every case.

Introduction / Context:
This is a remainder (congruence) problem. A number leaving the same remainder when divided by several moduli can be written as that common remainder plus a multiple of the least common multiple (LCM) of those moduli.

Given Data / Assumptions:

• Divisors: 12, 16, 18
• Common remainder: 5
• We seek the least positive such number.

Concept / Approach:
Let N be the number. Then N ≡ 5 (mod 12), N ≡ 5 (mod 16), and N ≡ 5 (mod 18). Therefore, N − 5 is a common multiple of 12, 16, and 18; hence N − 5 is a multiple of LCM(12, 16, 18).

Step-by-Step Solution:
Compute LCM: 12 = 2^2*3, 16 = 2^4, 18 = 2*3^2.LCM = 2^4 * 3^2 = 16 * 9 = 144.Thus N = 5 + 144t for integer t ≥ 0.Least positive solution corresponds to t = 1 ⇒ N = 149.

Verification / Alternative check:
149 ÷ 12 leaves remainder 5; 149 ÷ 16 leaves 5; 149 ÷ 18 leaves 5. All conditions hold.

Why Other Options Are Wrong:
139, 144, 154, and 161 do not equal 5 plus a multiple of 144, so they fail at least one of the remainder checks.

Common Pitfalls:
Using gcd instead of LCM; choosing t = 0 which gives 5 (too small compared to the options); or miscomputing the LCM exponents.

Final Answer:
149