Which of the following five digit numbers is exactly divisible by 99 without leaving any remainder?

Introduction / Context:
This question tests basic number theory and divisibility rules, specifically for the composite number 99. Since 99 is equal to 9 * 11, any number that is exactly divisible by 99 must satisfy the divisibility rules for both 9 and 11 at the same time. Understanding and applying these rules quickly is very useful in aptitude exams where calculations need to be done without a calculator.

Given Data / Assumptions:
- We have four candidate five digit numbers: 51579, 51557, 55036 and 49984. - We need to find which number is exactly divisible by 99. - A number divisible by 99 must be divisible by 9 and by 11.

Concept / Approach:
- 99 = 9 * 11, so divisibility by 99 implies divisibility by 9 and by 11. - Divisibility by 9: a number is divisible by 9 if the sum of its digits is divisible by 9. - Divisibility by 11: for a number, take the alternating sum of its digits (sum of digits in odd positions minus sum of digits in even positions in the same order). If the result is 0 or a multiple of 11, the number is divisible by 11.

Step-by-Step Solution:
Step 1: Check divisibility of 51579 by 9. Sum of digits = 5 + 1 + 5 + 7 + 9 = 27. Since 27 is divisible by 9, the number passes the rule for 9. Step 2: Check divisibility of 51579 by 11. Alternating sum = (5 + 5 + 9) - (1 + 7) = 19 - 8 = 11. Since 11 is a multiple of 11, 51579 is divisible by 11. Step 3: Since 51579 is divisible by both 9 and 11, it is divisible by 99. Step 4: For completeness, quickly see that the other numbers fail at least one rule. For example, 51557 has sum of digits 23 which is not divisible by 9. So it cannot be divisible by 99.

Verification / Alternative check:
We can also verify by direct division: 51579 / 99 gives an integer 521 with no remainder if we perform long division or use modular checks for 99 directly. Since both divisibility tests are satisfied, and quick checks reject the other options, further heavy computation is not needed.

Why Other Options Are Wrong:
Option B: 51557 has sum of digits 23 which is not divisible by 9, so it cannot be divisible by 99. Option C: 55036 has sum of digits 19 which is not divisible by 9. It fails the rule for 9, so it is not divisible by 99. Option D: 49984 has sum of digits 34 which is not divisible by 9, so it is not divisible by 99. Option E: 51599 has sum of digits 29 which is not divisible by 9, so this candidate also fails.

Common Pitfalls:
- Some learners check only one of the factors, usually 9, and forget to verify divisibility by 11. - Another common mistake is to compute the alternating sum for the 11 test incorrectly by mixing up positions or signs. - Trying full long division for every option wastes time in an exam. Using divisibility rules is much faster and less error prone.

Final Answer:
The only number that is divisible by both 9 and 11, and hence by 99, is 51579.