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
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
6*15 = 90 = 5*18
To find out which integers are also divisors of k, we need to find the common factors of 6 and 15.
First, let's list the factors of 6: 1, 2, 3, 6.
Next, let's list the factors of 15: 1, 3, 5, 15.
The common factors of 6 and 15 are 1 and 3.
Now, let's check which of the given integers (8, 12, 16, 18, 24) are also divisible by 1 and 3.
Starting with the first option, 8, we check if it is divisible by 1 and 3. Since 8 is not divisible by 3, it is not a factor/divisor of k.
Next, we check if 12 is divisible by 1 and 3. Since 12 is divisible by both 1 and 3, it is a factor/divisor of k.
Moving on to the next option, 16, we check if it is divisible by 1 and 3. Since 16 is not divisible by 3, it is not a factor/divisor of k.
Continuing with the next option, 18, we check if it is divisible by 1 and 3. Since 18 is divisible by both 1 and 3, it is a factor/divisor of k.
Finally, we check the last option, 24, if it is divisible by 1 and 3. Since 24 is divisible by both 1 and 3, it is a factor/divisor of k.
Therefore, the integers that are divisors of k are 12, 18, and 24. So, the answer is (B) 12, (D) 18, and (E) 24.