Last time, we found that if A, B, and C are vectors corresponding to cards that form a SET, then A + B + C = 0, mod 3. We also note that there nothing special about having 4 properties: we can let n be the number of properties, so each of our vectors are n-dimensional (meaning they have n components). Let’s explore our sum-to-0 condition a bit more, looking across all n ∈ ℕ.
First, let’s take a look at a small case: what happens when n is 2?
The 3 by 3 array above represents all 9 cards, or vectors with integer components modulo 3, in 2 dimensions. The three black dots can be represented by the vectors [02], [10], and [21]. These vectors indeed sum to 0, meaning they satisfy our conditions. What is interesting here is that we can take advantage of that fact that we are in mod 3. In other words, the translation (in one dimension) of any point by 3 is an identity transformation. This means the gray dot is equivalent to the black dot in the upper left corner, and we note that we can draw a line through our three dots! Thus, [02], [10], and [21] are collinear modulo 3 (even though the three black dots don’t look very collinear!).
Now, get some paper! Draw the array, and choose three vectors that do not sum to 0 (an example would be [01], [10], and [21]). Can you find a way to translate the points by multiples of 3 such that they are collinear?
Solution
It turns out that you can’t. In fact, for any vectors A, B, and C, we have that A + B + C ≡ 0 mod 3 if and only if they are collinear modulo 3.
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