Find the least number that must be added to 2497 so the resulting sum is exactly divisible by 5, 6, 4, and 3.

Introduction / Context:We are asked for the smallest nonnegative addition that makes a given number divisible by several integers at once. This is a classic application of least common multiple (LCM) and modular arithmetic.

Given Data / Assumptions:

• Starting number: 2497
• Divisors: 5, 6, 4, and 3
• We want 2497 + x to be divisible by each divisor.

Concept / Approach:Compute LCM(5, 6, 4, 3). The sum must be a multiple of this LCM. Then find the smallest x such that 2497 + x ≡ 0 (mod LCM).

Step-by-Step Solution:LCM(5, 6, 4, 3) = LCM(5, 6) = 30; LCM(30, 4) = 60; LCM(60, 3) = 60.Compute remainder of 2497 modulo 60: 2497 = 60*41 + 37, so 2497 ≡ 37 (mod 60).We need x such that 37 + x ≡ 0 (mod 60) ⇒ x ≡ 23 (mod 60).Smallest nonnegative x is 23.

Verification / Alternative check:2497 + 23 = 2520. Dividing: 2520/5 = 504, /6 = 420, /4 = 630, /3 = 840. All are integers.

Why Other Options Are Wrong:3, 13, 33, and 17 do not make 2497 + x a multiple of 60; they leave nonzero remainders modulo 60.

Common Pitfalls:Taking LCM incorrectly (e.g., multiplying all divisors without reduction) or using the wrong remainder when computing x modulo the LCM.

Final Answer:23