When 268 is added to 6K8, the result is 9P6. What is the least possible value of K and P for which 9P6 is divisible by 3?

Correct option is D

Given:

268 + 6K8 = 9P6

9P6 is divisible by 3

Concept Used:

Addition of three-digit numbers.

Divisibility rule for 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

Solution:

Perform the addition:

268

​6K8

9P6

Analyze the units column:

8 + 8 = 16. So, 6 is in the units place, and 1 is carried over.

Analyze the tens column:

6 + K + 1 (carry-over) = P

7 + K = P

Analyze the hundreds column:

2 + 6 = 8, but the result is 9, so there must have been a carry over of 1 from the tens column. Therefore K must be at least 3.

If K = 3, then 7+3 = 10, so P = 0, and a carry over of 1 occurs.

Apply the divisibility rule for 3 to 9P6:

9 + P + 6 = 15 + P

For 9P6 to be divisible by 3, 15 + P must be divisible by 3.

Find the least possible value of K:

From the addition, we know 7 + K = P.

If K = 3, then P = 10, but P is a single digit. Therefore we know there is a carry over from the tens column to the hundreds column.

Therefore, the addition in the tens column must result in a value of 10 or greater.

7+K = 10 + P.

K = 3+P.

Find the least possible value of P:

15 + P must be divisible by 3.

The smallest value of P that satisfies this is P = 0 (15 + 0 = 15, which is divisible by 3).

If P = 0, K = 3 + 0 = 3.

Verify the addition:

268

​638

906

906 is divisible by 3 because 9+0+6 = 15, which is divisible by 3.

The least possible value of K is 3, and the least possible value of P is 0.