How many positive integers#3

We can see that $1008$ is divisible by $3$ . Hence $3(336) + 5(0) = 1008$ is a trivial solution. $3(336-5) + 5(0 +3) = 1008$ is also a valid solution. We can continue to subtract 5 in the multiples of 3 and add 3 in the multiples of 5. Since the multiple of 5 will increase, we just need to count when the multiple of 3 gets negative and we can count the solutions. We can take the floor function of $\lfloor \frac{336}{5} \rfloor$ and we are done. This gives us $\boxed{67}$

From $3x+5y=1008$ , we note that the right-hand side is divisible by $3$ , therefore the left-hand side must also divisible by $3$ . This means that $y$ must be a multiple of $3$ . Let $y=3z$ , where $z$ is a positive integer. Then the equation becomes.

$\begin{aligned} 3x + 5(3z) & = 1008 \\ x + 5z & = 336 \\ \implies z & = \frac {336-x}5 \end{aligned}$

For $z$ to be an integer, $336-x$ must be divisible by $5$ , therefore $x$ is of the form $x=5n - 4$ , where $n$ is a positive integer. Then the largest $n$ is given by $336-(5n-4) \ge 0 \implies n = \dfrac {336+4}5 = 68$ . But when $n=68$ , $z=0 \implies y =0$ , which is unacceptable. Therefore there are $\boxed{67}$ positive integer pair $(x,y)$ solutions.