5. Positive integers $a, b$ and $c$ satisfy the equation $\frac{31}{72}=\frac{a}{8}+\frac{b}{9}-c$. What is the smallest possible value of $b$ ?
5. Since $\frac{31}{72}=\frac{a}{8}+\frac{b}{9}-c$, then multiplying both sides by 72 gives $31=9 a+8 b-72 c$.
Rearranging, we obtain $72 c-9 a=8 b-31$.
Since $a$ and $c$ are positive integers, then $72 c-9 a$ is an integer.
Since $72 c-9 a=9(8 c-a)$, then $72 c-9 a$ must be a multiple of 9 .
Since $72 c-9 a$ is a multiple of 9 , then $8 b-31=72 c-9 a$ must also be a multiple of 9 .
When $b=1,2,3,4$, we obtain $8 b-31=-23,-15,-7,1$, none of which is a multiple of 9 .
When $b=5$, we obtain $8 b-31=9$, which is a multiple of 9 .
Note further that if $b=5$, then the positive integers $c=1$ and $a=7$ gives $72 c-9 a=9=8 b-31$.
(There are other values of $c$ and $a$ that will give $72 c-9 a=9$, including $c=2$ and $a=15$.)
Thus, when $b=1,2,3,4$, there are no solutions, and when $b=5$, there is at least one solution.
Therefore, the smallest possible value of $b$ is 5 .
ANSWER: 5