In a divisibility test, the 7-digit number 78K928L is known to be divisible by 6. For which of the following pairs of digits (K, L) will this number be divisible by 6?

Introduction / Context:
This question tests the basic divisibility rules for the number 6 and how they apply to a multi-digit number containing unknown digits. Such problems are common in quantitative aptitude exams because they combine number properties with simple reasoning about digits. Understanding divisibility by 2 and 3 is very helpful for quickly checking whether large numbers are divisible by 6 without doing full division. Here, we must decide which choice of digits K and L makes the 7-digit number 78K928L divisible by 6.

Given Data / Assumptions:

• The number is 78K928L, where K and L are single digits from 0 to 9.
• The number is divisible by 6.
• We are given four possible pairs of values for (K, L) and must select the correct one.
• Standard divisibility rules for 2 and 3 apply.

Concept / Approach:
A number is divisible by 6 if and only if it is divisible by both 2 and 3. Divisibility by 2 requires that the last digit of the number is even. Divisibility by 3 requires that the sum of all digits of the number is a multiple of 3. We will apply these two checks for each pair of K and L values given in the options. Any pair that fails either of the two conditions cannot be correct. The pair that satisfies both conditions will be the required answer.

Step-by-Step Solution:
Step 1: Write the number as 7, 8, K, 9, 2, 8, L. Step 2: First enforce divisibility by 2. The last digit L must be even (0, 2, 4, 6, or 8). Step 3: Compute the sum of the fixed digits: 7 + 8 + 9 + 2 + 8 = 34. Step 4: For each option, add K and L to 34 and check if the total is a multiple of 3, while also making sure L is even. Step 5: Option (a) K=2, L=3: last digit 3 is odd, so the number is not divisible by 2, hence not divisible by 6. Step 6: Option (b) K=1, L=4: sum of digits is 34 + 1 + 4 = 39, which is a multiple of 3, and L=4 is even, so both divisibility rules are satisfied. Step 7: Option (c) K=1, L=2: sum is 34 + 1 + 2 = 37, not a multiple of 3, so it fails the divisibility by 3 test. Step 8: Option (d) K=3, L=3: again the last digit 3 is odd, so divisibility by 2 fails. Step 9: Therefore only option (b) gives a number divisible by 6.

Verification / Alternative check:
We can verify the result quickly by checking divisibility directly for the candidate in option (b). With K=1 and L=4, the number becomes 7819284. The last digit is 4, so the number is divisible by 2. The digit sum 7+8+1+9+2+8+4 equals 39, which is a multiple of 3, making the number divisible by 3. Since both conditions are met, the number is divisible by 6. This confirms that our earlier reasoning is correct and that no other option satisfies both rules simultaneously.

Why Other Options Are Wrong:
Option (a): L=3 makes the number odd, so it fails divisibility by 2.
Option (c): The digit sum becomes 37, which is not a multiple of 3, so it fails divisibility by 3.
Option (d): L=3 again makes the number odd, so it fails divisibility by 2.

Common Pitfalls:
A common mistake is to apply only the divisibility by 3 rule and forget that divisibility by 6 also requires divisibility by 2. Another error is computing the digit sum incorrectly or misreading the digits in the number. Some students also mistakenly check K and L independently instead of examining their combined effect on the digit sum. Careful and systematic checking of both conditions avoids these errors.

Final Answer:
The only pair of digits that makes 78K928L divisible by 6 is K = 1 and L = 4.