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

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

First, let's analyze the given information. We know that when 335 is added to 5A7, the result is 8B2.

Since 8B2 is divisible by 3, the sum of its digits must also be divisible by 3. Therefore, we can conclude that 8 + B + 2 is divisible by 3.

Now, let's consider the possible values of B. Since B is a digit, it can take values from 0 to 9.

To find the largest possible value of A, we need to find the largest value of B that satisfies the condition 8 + B + 2 is divisible by 3.

Let's test each value of B from 9 down to 0 and check if the sum is divisible by 3.

Starting with B = 9, we have 8 + 9 + 2 = 19, which is not divisible by 3.

Moving on to B = 8, we have 8 + 8 + 2 = 18, which is divisible by 3.

Therefore, the largest possible value of A is the digit that makes the sum divisible by 3 when B = 8.

To find this value, we subtract the sum of 8 and 2 from a multiple of 3. The largest multiple of 3 less than 18 is 15.

So, 15 - (8 + 2) = 5.

Therefore, the largest possible value of A is 5.