SOLUTION: If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following? (A)10 (B)12 (C)14 (D)16 (E)18 Thanks

If the positive integer is divisible by 6 and 8, then since 6 = 2×3 and
8 = 2×2×2 then it has 2 as a factor at least 3 times and 3 as a factor at least
one time, so the factors of any positive integer which we can be certain it is
divisible by must necessarily contain 2 as a factor three or fewer times and 3
no more than once.

(A) is not the correct answer.  It is not necessarily divisible by 10, since
10 = 2×5 and it does not necessarily have a factor of 5.

(C) is not the correct answer.  It is not necessarily divisible by 14, since
14 = 2×7 and it does not necessarily have a factor of 7.

(D) is not the correct answer.  It is not necessarily divisible by 16, since
16 = 2×2×2×2 and it does not necessarily have 2 as a factor 4 times.

(E) is not the correct answer.  It is not necessarily divisible by 18, since
18 = 2×3×3 and it does not necessarily have 3 as a factor 2 times.

That leaves (B) as the correct answer. It is necessarily divisible by 12, since
12 = 2×2×3 and therefore doesn't have 2 as a factor more than 3 times, and it
has 3 as a factor only once.

Edwin