If the number 517?324 is completely divisible by 3, then what is the smallest whole digit that can replace the question mark?

Introduction / Context:
This question asks us to find a missing digit in a large number so that the whole number becomes divisible by 3. The key tool is the divisibility rule for 3, which states that a number is divisible by 3 if the sum of its digits is a multiple of 3. This is a very common type of aptitude question.

Given Data / Assumptions:

- The number is 517?324. - The question mark represents a single digit from 0 to 9. - The full number must be divisible by 3. - We need the smallest such digit.

Concept / Approach:
According to the divisibility rule for 3, we only need to look at the sum of the digits of the number. If that sum is divisible by 3, then the number is divisible by 3. Therefore, we add up the known digits, represent the unknown digit by a variable, and then solve for the smallest digit that makes the total a multiple of 3.

Step-by-Step Solution:
Step 1: Let the unknown digit be x. The number is then 517x324. Step 2: Add the known digits: 5 + 1 + 7 + 3 + 2 + 4. Step 3: Compute this sum: 5 + 1 = 6, 6 + 7 = 13, 13 + 3 = 16, 16 + 2 = 18, 18 + 4 = 22. Step 4: The total sum of digits with x included is 22 + x. Step 5: For divisibility by 3, we require 22 + x to be a multiple of 3. Step 6: Since 22 leaves remainder 1 when divided by 3, we need x to leave remainder 2 when divided by 3, that is x ≡ 2 (mod 3). Step 7: The digits from 0 to 9 that satisfy x ≡ 2 (mod 3) are x = 2, 5 and 8. Step 8: The smallest such whole digit is 2.

Verification / Alternative check:
We can verify by substituting x = 2 into the number. The digits of 5172324 sum to 5 + 1 + 7 + 2 + 3 + 2 + 4 = 24, which is clearly divisible by 3. Therefore the number with x = 2 is divisible by 3. If we tried x = 1, the sum would be 23, not divisible by 3, so 1 would fail. This confirms that 2 is the smallest valid digit.

Why Other Options Are Wrong:
Option B (1) gives a sum of digits 23, option C (3) gives sum 25 and option D (7) gives sum 29. None of these totals is divisible by 3, so the corresponding numbers are not divisible by 3. Therefore they cannot be correct replacements for the missing digit.

Common Pitfalls:
Some students attempt to test each option by dividing the entire number by 3, which is time consuming and prone to error. Others may miscalculate the sum of digits or forget to find the smallest digit that works. Using modular reasoning and checking sums systematically is the quickest and safest method.

Final Answer:
The smallest whole digit that can replace the question mark is 2, which corresponds to option A.