Question

If 6 and 15 are both factors of k, then
which of the following integers are
also divisors of k?
(A) 8 (B) 12 (C) 16 (D) 18 (E) 24

Answer 1

6 = 2*3
15 = 3*5

testing:
so k = 2*3*3*5*c, where c is a prime number

8 = 2*2*2 , which will not divide into k
12 = 2*2*3 , no, not enough 2s in k
16 = 2*2*2*2, no
18 = 2*3*3 , no
24 = 2*2*2*3, no

e.g. suppose k = 2*3*5*29 = 870
870/6 = 145
870/15 = 58

870/8 = 108 3/4
870/12 = 72 1/2
870/16 = 54 3/8
870/18 = 48 1/3
870/24 = 36 1/4

none of them are, assuming that k does not contain any more of the factors shown in k = 2*3*5*c
of course the smallest value of k would be 30, (30 divides evenly by both 6 and 15)
and none of the given numbers would divided evenly into 30

Answer 2

6*15 = 90 = 5*18