1981 AHSME Problems/Problem 22

Problem

How many lines in a three dimensional rectangular coordinate system pass through four distinct points of the form $(i, j, k)$ , where $i$ , $j$ , and $k$ are positive integers not exceeding four?

$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 64\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 100$

Solution 1(casework)

Restating the problem, we seek all the lines that will pass through ( $i$ , $j$ , $k$ ), ( $i + a$ , $j + b$ , $k + c$ ), ( $i + 2a$ , $j + 2b$ , $k + 2c$ ), and ( $i + 3a$ , $j + 3b$ , $k + 3c$ ), such that $i,j,k$ are positive integers, $a,b,c$ are integers, and all of our points are between 1 and 4, inclusive. With this constraint in mind, we realize that for each coordinate, we have three choices:

Set $a/b/c$ to $0$ . This then allows us to set the corresponding $i,j,k$ to any number from $1$ to $4$ , inclusive.
Set $a/b/c$ to $1$ . This forces us to set the corresponding $i/j/k$ to $1$ .
Set $a/b/c$ to $-1$ . This forces us to set the corresponding $i/j/k$ to $4$ .

Note that options 2 and 3 will give us the same points if we mirror the assignments of each coordinate. Also note that we cannot set all three coordinates to not change, as that would be a point.
All of this gives us $6$ ways to assign each coordinate, for a total of $216$ . We then must subtract the ways to get a point ( $4$ ways per coordinate, for a total of $64$ ). This leaves us with $152$ . Finally, we divide by $2$ to account for mirror assignments giving us the same coordinate, for a final answer of $76$ .
(This was my first solution, apologies if it is bad).

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1981 AHSME Problems/Problem 22

Problem

Solution 1(casework)