Make 1056 divisible by 23 What is the least number that must be added to 1056 to obtain a number exactly divisible by 23?

Introduction / Context:
Finding the least addition to reach a multiple of a divisor is a standard modulus exercise. Compute the remainder and add the complement to the next multiple.

Given Data / Assumptions:

• N = 1056; divisor d = 23.
• We need k such that 1056 + k is a multiple of 23.
• k should be the smallest nonnegative integer that works.

Concept / Approach:
Let r = 1056 mod 23. If r = 0, k = 0. Otherwise, k = 23 − r. This ensures 1056 + k ≡ 0 (mod 23).

Step-by-Step Solution:
Compute 23 × 46 = 1058 ⇒ 1056 is 2 less than 1058.Therefore, 1056 mod 23 = 23 − 2 = 21, or simply note 1056 = 23×45 + 21.To reach the next multiple, add k = 23 − 21 = 2.Thus, the least number to add is 2.

Verification / Alternative check:
1056 + 2 = 1058 = 23 × 46, confirming exact divisibility.

Why Other Options Are Wrong:
3 or 21 overshoot the next multiple or correspond to different steps; 25 is unnecessary since 2 already works.

Common Pitfalls:
Confusing the remainder 21 with the needed addition; adding 21 instead of 2; not checking the very next multiple.

Final Answer:
2