Solution 820e0764-9b5b-4ba2-adcd-a3f32b034bf7
Points: 1 pt By definition \[ b = am \] multiply by $c$ \[ bc = a(mc) \] We see the integer $mc$ exists Final answer: a\mid bc
Points: 1 pt By definition \[ b = am \] multiply by $c$ \[ bc = a(mc) \] We see the integer $mc$ exists Final answer: a\mid bc