When 335 is added to 5A7, the result is 8B2. 8B2 is divisible by 3. What is the largest possible value of A ?Options8214

To find the largest possible value of A, we need to determine the range of values that A can take.

Given that 8B2 is divisible by 3, we know that the sum of its digits must be divisible by 3.

Let's break down the number 8B2 into its individual digits: 8, B, and 2.

We can represent the sum of these digits as: 8 + B + 2.

Since 335 is added to 5A7 to get 8B2, we can also represent the sum of the digits of 8B2 as: 3 + 3 + 5 + A + 7.

Setting these two representations equal to each other, we have:

8 + B + 2 = 3 + 3 + 5 + A + 7.

Simplifying this equation, we get:

B + 10 = 18 + A.

Rearranging the equation, we have:

B = 8 + A.

Since B is a digit, it must be between 0 and 9.

Substituting the possible values of B into the equation B = 8 + A, we can find the corresponding values of A.

If B = 0, then A = 8 + 0 = 8. If B = 1, then A = 8 + 1 = 9. If B = 2, then A = 8 + 2 = 10. If B = 3, then A = 8 + 3 = 11. If B = 4, then A = 8 + 4 = 12. If B = 5, then A = 8 + 5 = 13. If B = 6, then A = 8 + 6 = 14. If B = 7, then A = 8 + 7 = 15. If B = 8, then A = 8 + 8 = 16. If B = 9, then A = 8 + 9 = 17.

Out of these possible values of A, the largest one is 17.

Therefore, the largest possible value of A is 17.