In divisibility tests for aptitude simplification, determine which of the following numbers is completely divisible by 99 (that is, divisible by both 9 and 11).

Introduction / Context:
This question tests your understanding of divisibility rules, particularly for the composite number 99. Since 99 = 9 * 11, a number is divisible by 99 exactly when it is divisible by both 9 and 11. Using the divisibility tests for 9 and 11 allows us to check each option quickly without performing full long division. This is a common skill tested in quantitative aptitude exams.

Given Data / Assumptions:

• We are given four candidate numbers: 57717, 57627, 55162, and 56982.
• We must find which number is divisible by 99.
• A number divisible by 99 must be divisible by both 9 and 11.
• Standard divisibility tests for 9 and 11 are used.

Concept / Approach:
The divisibility test for 9: a number is divisible by 9 if the sum of its digits is a multiple of 9. The test for 11: a number is divisible by 11 if the difference between the sum of the digits in odd positions and the sum of the digits in even positions (from left) is a multiple of 11 (including 0). We apply both tests to each option. A number that passes both tests is divisible by 99.

Step-by-Step Solution:
1) Consider 57717. Sum of digits = 5 + 7 + 7 + 1 + 7 = 27, which is a multiple of 9, so it is divisible by 9. 2) For the 11-test, assign positions from the left: digits are 5(1), 7(2), 7(3), 1(4), 7(5). 3) Sum of digits in odd positions (1, 3, 5) = 5 + 7 + 7 = 19. Sum in even positions (2, 4) = 7 + 1 = 8. Difference = 19 − 8 = 11, which is a multiple of 11, so 57717 is divisible by 11. 4) Therefore 57717 is divisible by both 9 and 11, so it is divisible by 99. 5) Quickly check another option, 57627. Sum of digits = 5 + 7 + 6 + 2 + 7 = 27, so it is divisible by 9. 6) For 11-test on 57627: odd positions (1, 3, 5) → 5 + 6 + 7 = 18; even positions (2, 4) → 7 + 2 = 9; difference = 18 − 9 = 9, not a multiple of 11, so 57627 is not divisible by 11 and hence not by 99.

Verification / Alternative check:
We can confirm by dividing 57717 by 99. First divide 57717 by 9: 57717 / 9 = 6413. Then check 6413 / 11 = 583, which is an integer. Therefore 57717 = 99 * 583, confirming that it is exactly divisible by 99. Similar checks for other options would either fail at 9 or at 11, confirming that they are not divisible by 99.

Why Other Options Are Wrong:
For 57627, although the digit sum is 27 (divisible by 9), the 11-test fails, so it is not divisible by 99. The number 55162 has digit sum 5 + 5 + 1 + 6 + 2 = 19, which is not a multiple of 9, so it fails immediately. The number 56982 has digit sum 5 + 6 + 9 + 8 + 2 = 30, which is not a multiple of 9 either. Therefore, options b, c, and d are not divisible by 99. Option e, 'None of these', is wrong because we have already found a valid number, 57717.

Common Pitfalls:
Students sometimes misapply the 11-test by adding all digits instead of forming the alternating sum, or they miscount positions from the right instead of the left. It is also easy to miscalculate digit sums under exam pressure. Carefully assigning positions and computing sums helps avoid such mistakes. Always remember that divisibility by 99 requires divisibility by both 9 and 11, not just one of them.

Final Answer:
The number that is completely divisible by 99 is 57717.