Prove that, if ABA is divisible by 7 (with no remaining), then BAB is also divisible by 7 (with no remaining) as well.
Example 1; 161 = 7x23, 616 = 7x88, both 161 and 616 are divisible by 7 (with no remaining).
Example 2; 252 = 7x36, 525 = 7x75, both 252 and 525 are divisible by 7 (with no remaining).