If 499A7B2 is divisible by 72, then find maximum value of (A+B)?

Options

A. 10
B. 14
C. 9
D. 8

Answer

To solve the problem of finding the maximum value of \( A + B \) such that the number \( 499A7B2 \) is divisible by 72, we need to ensure that it meets the divisibility rules for both 8 and 9, since \( 72 = 8 \times 9 \).

### Step-by-Step Solution:

**Step 1: Check divisibility by 8**
To check if a number is divisible by 8, we look at the last three digits of the number. In this case, the last three digits are \( 7B2 \).

We need \( 7B2 \) to be divisible by 8. We can test values of \( B \) from 0 to 9:

- If \( B = 0 \): \( 702 \div 8 = 87.75 \) (not divisible)
- If \( B = 1 \): \( 712 \div 8 = 89 \) (divisible)
- If \( B = 2 \): \( 722 \div 8 = 90.25 \) (not divisible)
- If \( B = 3 \): \( 732 \div 8 = 91.5 \) (not divisible)
- If \( B = 4 \): \( 742 \div 8 = 92.75 \) (not divisible)
- If \( B = 5 \): \( 752 \div 8 = 94 \) (divisible)
- If \( B = 6 \): \( 762 \div 8 = 95.25 \) (not divisible)
- If \( B = 7 \): \( 772 \div 8 = 96.5 \) (not divisible)
- If \( B = 8 \): \( 782 \div 8 = 97.75 \) (not divisible)
- If \( B = 9 \): \( 792 \div 8 = 99 \) (divisible)

Thus, \( B \) can be \( 1, 5, \) or \( 9 \).

**Step 2: Check divisibility by 9**
For a number to be divisible by 9, the sum of its digits must be a multiple of 9. The sum of the digits of \( 499A7B2 \) is:

\[ 4 + 9 + 9 + A + 7 + B + 2 = 31 + A + B \]

We need \( 31 + A + B \) to be a multiple of 9.

**Step 3: Evaluate possible values of \( A + B \)**
Now we will check the values of \( B \) we found earlier and calculate \( A + B \):

1. **For \( B = 1 \)**:
\[ 31 + A + 1 = 32 + A \]
We need \( 32 + A \equiv 0 \mod 9 \). The remainder when 32 is divided by 9 is 5. Thus, \( A \equiv 4 \mod 9 \). The possible values for \( A \) are 4.

So, \( A + B = 4 + 1 = 5 \).

2. **For \( B = 5 \)**:
\[ 31 + A + 5 = 36 + A \]
We need \( 36 + A \equiv 0 \mod 9 \). Since 36 is already a multiple of 9, \( A \) can be 0, 9, etc. The maximum value of \( A \) is 9.

So, \( A + B = 9 + 5 = 14 \).

3. **For \( B = 9 \)**:
\[ 31 + A + 9 = 40 + A \]
We need \( 40 + A \equiv 0 \mod 9 \). The remainder when 40 is divided by 9 is 4. Thus, \( A \equiv 5 \mod 9 \). The possible value for \( A \) is 5.

So, \( A + B = 5 + 9 = 14 \).

### Conclusion
The maximum value of \( A + B \) is \( 14 \).