Dribbling The Numbers

2a + 3b = 92

Evidently b is even. Let b= 2k.

a + 3k = 46

So k can take any value from 0 to 15 , to be non-negative itself, and yield non-negative a.

16 possibilities.

Let the number of two-pointers be $n$ , and the number of three-pointers be $m$ . Thus,

$2n+3m=92$

$\implies 92-2n=2(46-2n)=3m$

$\implies m = \frac{1}{3}(92-2n)$

Also,

$n=\frac{1}{2}(92-3m)=46-\frac{3}{2}m$

We see that $m$ must be even. Begin by letting $m=0$ :

$m=0\implies 2n=92\implies n=46$

Our first solution $(n, m)$ is thus $(46, 0)$ .

Now, consider another possible number of three-pointers $m'$ . Let $m'=m+2$ . Assume that there is a solution $(n', m')$ . Then,

$2n'+3(m+2)=92\implies 2n'+3m=86$ $n'=\frac{1}{2}(86-3m)=43-\frac{3}{2}m=n-3$

Thus, any ordered pair $(n', m')$ such that $n'=n-3$ and $m'=m+2$ , is a solution.

As $46=3(15)+1$ , there are $\large{15+1= \boxed{16}}$ possible combinations of two-pointers and three-pointers, when a team has won $92$ points.