A four-digit number has the digit form "a4b2." It is divisible by 3. What are the possible values of a + b? Please include your calculation.

To determine if a number is divisible by 3, we need to find the sum of its digits and see if that sum is divisible by 3.

In this case, the number has the digit form "a4b2." So, the sum of the digits is a + 4 + b + 2 = a + b + 6.

For the number to be divisible by 3, the sum of the digits must be divisible by 3. Therefore, a + b + 6 must be divisible by 3.

We can rewrite this as (a + b) + 6 = 3n, where n is an integer.

Subtracting 6 from both sides gives us a + b = 3n - 6.

Since a and b are single digits, the possible values of a + b are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Therefore, the possible values of a + b are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Calculation:

The sum of the digits of the number is: a + 4 + b + 2 = a + b + 6

For the number to be divisible by 3, the sum of its digits must be divisible by 3. Therefore, a + b + 6 must be divisible by 3.

The possible values of a + b that make a + b + 6 divisible by 3 are:
- 3
- 6
- 9
- 12
- 15
- 18

Possible Values of a and b:

Since a and b are digits, their values can range from 0 to 9.

For a + b = 3:
- a = 0, b = 3
- a = 1, b = 2

For a + b = 6:
- a = 2, b = 4
- a = 3, b = 3
- a = 4, b = 2

For a + b = 9:
- a = 3, b = 6
- a = 4, b = 5
- a = 5, b = 4
- a = 6, b = 3

For a + b = 12:
- a = 4, b = 8
- a = 5, b = 7
- a = 6, b = 6
- a = 7, b = 5
- a = 8, b = 4

For a + b = 15:
- a = 5, b = 10 (not possible since b is a digit)

For a + b = 18:
- a = 6, b = 12 (not possible since b is a digit)

Therefore, the possible values of a and b are:

- a = 0, b = 3
- a = 1, b = 2
- a = 2, b = 4
- a = 3, b = 3
- a = 4, b = 2
- a = 3, b = 6
- a = 4, b = 5
- a = 5, b = 4
- a = 6, b = 3
- a = 4, b = 8
- a = 5, b = 7
- a = 6, b = 6
- a = 7, b = 5
- a = 8, b = 4