If a 679b is a five digit number that is divisible by 72, then a+b is equal to

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the five-digit number \( 679b \) is divisible by 72. We will use the divisibility rules for both 8 and 9, since \( 72 = 8 \times 9 \).

### Step-by-Step Solution:

1. **Understand the Divisibility Rules**:
- A number is divisible by **8** if its last three digits form a number that is divisible by 8.
- A number is divisible by **9** if the sum of its digits is divisible by 9.

2. **Apply the Divisibility Rule for 8**:
- The last three digits of our number are \( 79b \).
- We need to find \( b \) such that \( 79b \) is divisible by 8.
- We can check the values of \( b \) from 0 to 9:
- \( 790 \div 8 = 98.75 \) (not divisible)
- \( 791 \div 8 = 98.875 \) (not divisible)
- \( 792 \div 8 = 99 \) (divisible)
- \( 793 \div 8 = 99.125 \) (not divisible)
- \( 794 \div 8 = 99.25 \) (not divisible)
- \( 795 \div 8 = 99.375 \) (not divisible)
- \( 796 \div 8 = 99.5 \) (not divisible)
- \( 797 \div 8 = 99.625 \) (not divisible)
- \( 798 \div 8 = 99.75 \) (not divisible)
- \( 799 \div 8 = 99.875 \) (not divisible)
- The only value of \( b \) that works is \( b = 2 \).

3. **Apply the Divisibility Rule for 9**:
- Now, we need to check the sum of the digits \( a + 6 + 7 + 9 + b \).
- Substituting \( b = 2 \), we have:
\[ a + 6 + 7 + 9 + 2 = a + 24 \]
- We need \( a + 24 \) to be divisible by 9.

4. **Find Possible Values for \( a \)**:
- The multiples of 9 around 24 are 18, 27, and 36.
- Since \( a \) must be a single digit (0-9), we can only consider \( 27 \) as the next multiple of 9 after 24.
- Setting \( a + 24 = 27 \):
\[ a = 27 - 24 = 3 \]

5. **Calculate \( a + b \)**:
- Now we have \( a = 3 \) and \( b = 2 \).
- Therefore, \( a + b = 3 + 2 = 5 \).

### Final Answer:
\[ a + b = 5 \]