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
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
RedeemTarget Test Prep · GMAT
Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

with Chris Peckover

with Logan Thompson
Complete access from day one. Study on your schedule.
Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.
DanaJ, thanks for the reminder! Once in while too, I get caught in such ques.DanaJ wrote: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.
New here Create free account