Least number with a fixed remainder: Find the smallest number that leaves remainder 2 when divided by each of 12, 15, 20, and 24.

Introduction / Context:Numbers leaving the same remainder upon division by several divisors are handled using the Least Common Multiple (LCM). The pattern is N ≡ r (mod each divisor), hence N − r is a common multiple of all divisors.

Given Data / Assumptions:

• Divisors: 12, 15, 20, 24
• Common remainder: r = 2
• We seek the least positive N such that N ≡ 2 for all divisors.

Concept / Approach:Compute LCM of the divisors, then set N = LCM + r for the least positive solution. In general, all solutions are N = LCM * k + r, k ≥ 1 integer, but k = 1 gives the smallest positive answer beyond r itself (which will not satisfy divisibility here).

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• 120 is divisible by the moduli (remainder 0, not 2).
• 124, 126 do not leave remainder 2 with all the given divisors simultaneously.

Common Pitfalls:

• Mistaking LCM itself as the answer rather than adding the fixed remainder.

Final Answer: