If $a679b$ is the decimal expansion of a number in base $10$, such that it is divisible by  $72$, determine $a,b$.
To determine the values of \(a\) and \(b\) such that the number \(a679b\) is divisible by \(72\), we need to check the divisibility rules for both \(8\) and \(9\), since \(72 = 8 \times 9\).

1. **Divisibility by 8**:
   A number is divisible by \(8\) if its last three digits form a number that is divisible by \(8\). Therefore, we need to check the number \(79b\).

   - We test each digit \(b\) from \(0\) to \(9\) to find which makes \(79b\) divisible by \(8\):
     \[
     \begin{aligned}
     790 &\div 8 = 98.75 \quad (\text{not divisible}) \\
     791 &\div 8 = 98.875 \quad (\text{not divisible}) \\
     792 &\div 8 = 99 \quad (\text{divisible}) \\
     793 &\div 8 = 99.125 \quad (\text{not divisible}) \\
     794 &\div 8 = 99.25 \quad (\text{not divisible}) \\
     795 &\div 8 = 99.375 \quad (\text{not divisible}) \\
     796 &\div 8 = 99.5 \quad (\text{not divisible}) \\
     797 &\div 8 = 99.625 \quad (\text{not divisible}) \\
     798 &\div 8 = 99.75 \quad (\text{not divisible}) \\
     799 &\div 8 = 99.875 \quad (\text{not divisible}) \\
     \end{aligned}
     \]
   - The only value for \(b\) that makes \(79b\) divisible by \(8\) is \(b = 2\).

2. **Divisibility by 9**:
   A number is divisible by \(9\) if the sum of its digits is divisible by \(9\). Therefore, we need to check the sum of the digits of \(a6792\):
   \[
   a + 6 + 7 + 9 + 2 = a + 24
   \]
   - We need \(a + 24\) to be divisible by \(9\). Since \(a\) is a single digit, we test values from \(0\) to \(9\):
     \[
     \begin{aligned}
     a + 24 &= 27 \quad (\text{divisible by 9}) \\
     \end{aligned}
     \]
   - Solving for \(a\):
     \[
     a + 24 = 27 \implies a = 27 - 24 = 3
     \]

Thus, the values of \(a\) and \(b\) that make \(a679b\) divisible by \(72\) are \(a = 3\) and \(b = 2\).

The final answer is \( \boxed{ a = 3 } \) and \(b = 2\).