January 6, 2023
You and a friend are shooting some hoops at your local basketball court when she issues a challenge: She will name a number, which we’ll call N. Your goal is to score exactly N points in as many ways as possible using only 2-point and 3-point shots. The order of your shots does not matter
For example, there are two ways you could score N = 8 points: four 2-pointers or two 3-pointers and one 2-pointer.
Your apparently sadistic friend chooses 60 for the value of N. You try to negotiate this number down, but to no avail. However, she says you are welcome to pick an even larger value of N. Does there exist an integer N greater than 60 such that there are fewer ways to score N points than there are ways to score 60 points?
\(N = 61\)
Explanation:
Let \(x\) be the number of three pointers you need to score and \(y\) be the number of two pointers. Please note that \(x\) and \(y\) must be non-negative integers. Then we have:
\begin{align} 3x + 2y &= N \\ \end{align}
The above is an example of a linear Diophantine equation. The key to solving those equations when they are linear is to make a table and think critically about the conditions that satisfy those equations. For the one that we have, we note that \(2y\) is always even. If \(N\) is an even number like 60, then \(3x\) must be even as well to make the sum even. Since 3 is odd, that means \(x\) must be even, and so our table looks like the following for \(N = 60\):
| \(x\) | \(y\) |
|---|---|
| 0 | 30 |
| 2 | 27 |
| 4 | 24 |
| 6 | 21 |
| 8 | 18 |
| \(\dots\) | \(\dots\) |
| 20 | 0 |
Therefore, for \(N = 60\), there are 11 ways to score it with 3-pointers and 2-pointers with \(x = 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20\).
The key here is to realize that we cannot score 1-pointers, and so if we have \(N = 61\), then \(x = 20\) is out of the question as we would have 60 points and no way to make an additional 1 pointer. Also, here since \(N\) is odd, we need to have \(x\) being odd as well. This means, by similar analysis for our linear Diophantine equation, that \(x = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19\) for only 10 ways, which is less than 11.
It is also clear that for any \(N > 61\), we would have at least 11 ways to score that many points. For example, for \(N = 62\), the cases are the same as that for \(N = 60\) with the number of 3-pointers. For \(N = 63\), since it is divisible by 3, we have an extra case with \(x = 21\) and \(y = 0\). Any integer greater than 63 would introduce more cases for \(x\) and \(y\).
Therefore, \(N = 61\) is the only answer!