If an integer n is divisible by both 6 and 8, then it must also be divisible by which of the following?
Though I got the correct ans by brute force. Just wondering which math rule could be applied here.
The way I figure it is taking the product of the prime factors which are not common among the two.
prod. of prime factors not common = 2*2*3=12
I like this explanation. If it's divisible by 6, it must have every prime 6 has, and the same thing is true for 8. So there must be 2,2,3. Even if you don't find the lowest common multiple, as in above, you should still just look for the answers with just some 2's and one 3. 12 is the only one.
Note that 18 doesn't work because there's an extra 3 in it.