Do u have a problem that went with this?
The answer to this question is A prime number. A prime number has only two factors one an itself. Opposite to composite which has more than two. Example!
Prime:2,3,5,7
Composite:4,6,8,9
You're trying to find constants

such that

. Equivalently, you're looking for the least-square solution to the following matrix equation.

To solve

, multiply both sides by the transpose of

, which introduces an invertible square matrix on the LHS.

Computing this, you'd find that

which means the first choice is correct.
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