Determine the number of pairs $(x, y)$ of positive integers for which $0<x<y$ and $2 x+3 y=80$.
Since $02 x+3 x=5 x$.

Since $2 x+3 y=80$ and $2 x+3 y>5 x$, then $80>5 x$ or $16>x$.

Since $2 x+3 y=80$, then $2 x=80-3 y$. Since $2 x$ is even, then $80-3 y$ is even. Since 80 is even, then $3 y$ is even, and so $y$ must be even.

Set $y=2 Y$ for some positive integer $Y$.

Then $2 x+3 y=80$ becomes $2 x+6 Y=80$ or $x+3 Y=40$ or $x=40-3 Y$.

Since $x24$ or $Y>8$.

Since $x>0$, then $3 Y<40$ or $Y<13 \frac{1}{3}$.

When $Y=9$ (that is, $y=18$ ), $x=13$.

When $Y=10$ (that is, $y=20$ ), $x=10$.

When $Y=11$ (that is, $y=22$ ), $x=7$.

When $Y=12$ (that is, $y=24$ ), $x=4$.

When $Y=13$ (that is, $y=26$ ), $x=1$.

Therefore, there are 5 pairs that satisfy the given conditions: $(x, y)=(13,18),(10,20),(7,22)$, $(4,24),(1,26)$.

ANSWER: 5