How many pairs of positive integers $(n, m)$ are there such that $2n+3m=2015$?

Question

I know that $m$ must be odd and $m\le671$. Also, $n\le1006$. I can't go any further, any help?

Answer 1

Since $2$ and $3$ are coprime, every integer solution $(a,b)$ of $2a+3b=2015$ is of the form $a=1+3k,b=671-2k$ for some integer $k$. In order that both $a$ and $b$ are positive, $k$ must lie between $0$ and $335$, hence there are $\color{red}{336}$ positive integer solutions.

Answer 2

Hint: Any odd $m$ in the range will do it: express $n$ from the equation.

Answer 3

$$\begin{array} {|cc|} \hline 1 & 2014 \\ {\color{red}{ 2 }} & {\color{blue}{ 2013 }} \\ 3 & 2012 \\ {\color{red}{ 4 }} & 2011 \\ 5 & {\color{blue}{ 2010 }} \\ {\color{red}{ 6 }} & 2009 \\ 7 & 2008 \\ {\color{red}{ 8 }} & {\color{blue}{ 2007 }} \\ 9 & 2006 \\ {\color{red}{ 10 }} & 2005 \\ 11 & {\color{blue}{ 2004 }} \\ {\color{red}{ 12 }} & 2003 \\ 13 & 2002 \\ {\color{red}{ 14 }} & {\color{blue}{ 2001 }} \\ 15 & 2000 \\ \vdots & \vdots \end{array}$$

The ones where both red and blue are highlighted, $2$, $8$, $14$,$\dots$, which are $2$ more than a multiple of $6$. So it's $\left \lfloor \frac {2015 - 2}{6} \right \rfloor + 1 = 336$.