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
OA:B
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
OA:B
Well, the trick here is to know what exactly divisibility tells you. If it's divisible by both 6 and 8, then it's divisible by their LCM, 24.
Find the LCM this way: 6 = 2*3
8 = 2^3
Take the highest power of every prime number and multiply it to get LCM = 3*(2^3) = 24.
If it's divisible by 24, then it's divisible by any divisor of 24: 2, 4, 6, 8, 12, 24....
A. 10 is 2*5 - there's no 5 in 24
B. 12 is 2*6 - 24 is divisible by 12. On test day, you stop here and ignore the other answers. I'll just keep on going for the sake of exercise.
C. 14 is 2*7 - you don't have 7 in 24
D. 16 is a trap, actually. Most people would think that if a number is divisible by 6 and 8, then it's divisible by their product, 48, which is also divisible by 16. HOWEVER, it's not their product that divides the number in question, but rather their LCM. Don't make that mistake on test day!
E. 18 is not a divisor of 24 either.
