For any integer between 100 and 1000, if you subtract the sum of its digits from the integer itself, the resulting number is always divisible by which of the following numbers?

Introduction / Context:

This number system question tests your understanding of divisibility rules and how numbers can be expressed in terms of their digits. It is a favourite type in aptitude and competitive exams because it checks both conceptual understanding and familiarity with properties of the number 9.

Given Data / Assumptions:

• We consider any integer N between 100 and 1000 (a three digit number).
• The sum of its digits is subtracted from N.
• We must determine a number from the options that always divides the result exactly.
• The usual base 10 representation of numbers is assumed.

Concept / Approach:

Write a three digit number N in terms of its digits: N = 100a + 10b + c, where a, b, and c are digits and a is not zero. The sum of the digits is a + b + c. Compute N minus (a + b + c) and look for a common factor. The property that numbers of the form 9k are divisible by 9 will give the answer.

Step-by-Step Solution:

Verification / Alternative check:

Test with a sample, for example N = 572. Sum of digits is 5 + 7 + 2 = 14. Then N - S = 572 - 14 = 558. Since 558 divided by 9 equals 62, 9 divides the result. Other three digit numbers will behave similarly because the algebraic form is general.

Why Other Options Are Wrong:

The values 2, 5, 6, and 7 do not always divide numbers of the form 9(11a + b). Only 9 is guaranteed, because 9 is the explicit factor. For certain numbers, factors like 2 or 5 may divide the result, but this is not guaranteed for every three digit number.

Common Pitfalls:

Learners sometimes guess 3, since 9 is a multiple of 3, but the option here highlights 9 directly. Another common mistake is to test only one or two numbers and then jump to a conclusion instead of using an algebraic argument.

Final Answer:

The resulting number is always divisible by 9.