# Modular Linear Congruences Puzzle

I'm learning about solving systems of modular linear congruences in my discrete mathematics class. Recently, my teacher posed a puzzle that I can't seem to solve:

These eight small-number triples are not random:

[[1 1 3] [1 1 4] [1 4 3] [1 4 4] [2 1 3] [2 1 4] [2 4 3] [2 4 4]]

They have something to do with the product of the first three odd primes and the fourth power of two.

Find the connection.

From what I can tell, the triples are the cartesian products of [1 2], [1 4], and [3 4]. These add up to the first three odd primes like the teacher wanted. I still can't find a link between the triples and the fourth power of two though. My teacher said it has something to do with modular linear congruences. What am I missing?

This is an example of modular linear congruences:

$x \equiv_7 0$

$x \equiv_{11} 8$

$x \equiv_{13} 12$

Solution: $x \equiv_{1001} 987$

I think I may have found a connection. If you use the Chinese Remainder Theorem with 3, 5, and 7 as the mods, and testing it with each set of triples as remainders, you get this list:

74,59,11,101,4,94,46,31

putting them in ascending order: 4, 11, 31, 46, 59, 74, 94, 101

Now if you mod the square of each of those numbers with the product of the first three primes, you get 16.

For example:

16 mod 105 = 16
121 mod 105 = 16
961 mod 105 = 16


etc.