Number System — Divisibility Adjustment by 33 What is the smallest whole number that must be added to 231228 so that the result is exactly divisible by 33?

Introduction / Context:This problem checks quick divisibility using remainders (modular arithmetic). We are asked to find the least non-negative integer that, when added to 231228, produces a multiple of 33. Such questions are common in aptitude tests because they reward number sense and speed without heavy computation.

Given Data / Assumptions:

• Dividend candidate: 231228.
• Divisor: 33.
• We want the smallest added value k such that 231228 + k is divisible by 33.
• k must be a whole number (0, 1, 2, ...).

Concept / Approach:Use the remainder idea: if N % 33 = r, then N + (33 - r) is the next multiple of 33, unless r = 0 (in which case k = 0). The goal is to compute the remainder efficiently and adjust to the next multiple.

Step-by-Step Solution:1) Compute r = 231228 % 33.2) Divide 231228 by 33 to find the remainder. The remainder is 30.3) The next multiple is obtained by adding 33 - r = 33 - 30 = 3.4) Therefore k = 3 is the least number to add.

Verification / Alternative check:Compute (231228 + 3) / 33 = 231231 / 33 = 701 (an integer), confirming exact divisibility. Any smaller k (0, 1, 2) would keep a non-zero remainder.

Why Other Options Are Wrong:1 and 2 leave remainders 31 and 32 respectively; 4 overshoots the next multiple; 5 also overshoots and is not minimal.

Common Pitfalls:Adding until you find a multiple by trial rather than computing the remainder; confusing subtract with add when asked for the next multiple (upward adjustment is required here).

Final Answer:3