# Suppose A, B, and C are positive integers such that

Suppose A, B, and C are positive integers such that $\frac{32}{5}=A+\frac{1}{B+\frac{1}{c+1}}$. Then the values of $3A + 2B + 5C$ equals?

$\frac{32}{5}=6.4$ Then I know that $A=6$, the part I get stuck on is trying to find $B$ and $C$.

You have correctly concluded that $A=6$. Then we have, $$\frac {32}{5}-6 = \frac {2}{5} = \frac {1}{B + \frac {1}{C+1}} = \frac {C+1}{BC+B+1}$$ $$\Rightarrow C = \frac {2B-3}{5-2B}$$

Now we can easily see that for $C$ to be positive, there are two possibilities:

Case $1$: Both numerator and denominator are positive $\implies 5 > 2B > 3$. We can see that $B=2$ is the only satisfying integer.

Case $2$: Both numerator and denominator are negative $\implies 2B < 3 \text { and } 2B > 5$ which is not possible.

Thus, $C=1$ and the result follows. Hope it helps.

• I get it know, so essentially we solve for C and check our parameters to make sure they are positive integers. Thank you – Killercamin Feb 14 '17 at 17:58

If a number $\alpha$ equals an integer $N$ plus the reciprocal of a number $\beta>1$, then certainly $N=\lfloor \alpha\rfloor$ and after that $\beta=\frac1{\alpha-N}$. So here $$\frac{32}5=A+\frac1{B+\frac1{C+1}}$$ implies $A=\lfloor\frac{32}5\rfloor = 6$ and then $$\frac1{\frac{32}5-A}=\frac52=B+\frac1{C+1}$$ implies $B=\lfloor \frac 52\rfloor =2$. After that $$\frac1{\frac 52-B}=2=C+1,$$ so $C=1$.