Find the smallest number that must be added to 756896 so that the resulting sum is exactly divisible by 11.

Introduction / Context:
This problem focuses on divisibility by 11, a common topic in number system questions in aptitude exams. Instead of performing long division directly, it is usually more efficient to use the divisibility rule for 11 or modular arithmetic. You must find the smallest non negative integer that, when added to 756896, gives a number divisible by 11.

Given Data / Assumptions:

• The starting number is 756896.
• We want to add the smallest possible number k so that 756896 + k is divisible by 11.
• The answer should be a non negative integer, and usually a small one.
• All calculations use standard integer arithmetic.

Concept / Approach:
The key concept here is working with remainders modulo 11. If a number leaves a remainder r when divided by 11, then we need to add (11 - r) to reach the next multiple of 11, unless the remainder is 0. There is also a digit based divisibility rule for 11 that uses alternating sums of digits, but calculating the remainder directly is often simpler when you are comfortable with modular arithmetic.

Step-by-Step Solution:
Step 1: Let k be the smallest number to add to 756896 so that 756896 + k is divisible by 11. Step 2: First find the remainder when 756896 is divided by 11. Step 3: Compute 756896 mod 11. Using division or a calculator, we find that 756896 leaves a remainder of 8 when divided by 11. Step 4: We want 756896 + k to have remainder 0 modulo 11, which means (8 + k) mod 11 must be 0. Step 5: So k must satisfy 8 + k = 11 or 8 + k = 22 and so on. The smallest positive solution is k = 3. Step 6: Check: 756896 + 3 = 756899. Step 7: Verify that 756899 divided by 11 has no remainder. Since 756896 had remainder 8, adding 3 increases the remainder to 11, which is equivalent to 0, confirming divisibility.

Verification / Alternative check:
As an alternative method, you can use the divisibility rule for 11 which states that the difference between the sum of digits in odd positions and the sum of digits in even positions should be a multiple of 11. However, because the number is large, modular arithmetic is more straightforward. Another verification is to try adding small values 1, 2, 3 in order and check remainders. Adding 1 gives remainder 9, adding 2 gives remainder 10, and adding 3 gives remainder 0, which confirms that 3 is indeed the smallest required addition.

Why Other Options Are Wrong:
If we add 1, the new remainder is 9, so the sum is not divisible by 11. Adding 2 results in remainder 10. Adding 4 yields remainder 1, and adding 5 yields remainder 2. All of these still leave a non zero remainder when divided by 11. Only adding 3 converts the remainder from 8 to 0, making 756899 the nearest multiple of 11 greater than 756896.

Common Pitfalls:
A common mistake is to attempt full division by 11, which is slower and more error prone, especially under exam pressure. Another pitfall is confusing the direction of the adjustment and trying to subtract a number instead of adding one. Some students also mistakenly assume that the added number must be a multiple of 11 itself. Remember that we only care about changing the remainder to zero, which is often achievable with a small adjustment like this case. Practicing remainder based thinking greatly simplifies such questions.

Final Answer:
The smallest number that must be added to 756896 to obtain a multiple of 11 is 3.