Prove if c divides a and c divides b, then c divides any combination an+bm, where n and m are integers.
Since c divides both a and b then there must exist integers j and t such that a=cj and b=ct.
So by substitution into an+bm we get
an+bm = (cj)n+(ct)m = c (jn + tm). So c divides (an + bm) ~~~~
Thus, if c divides a and b, then c also divides any multiple of a and b.
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