andih wrote:
Hello,
quick question, I got the answer right, though I am having some problems with the "plug in numbers" approach here. Can anybody help real quick with an example?
Thanks in advance!
If the positive integer x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following?I. 8
II. 12
III 18
A. II only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III
First of all notice that we are asked \(xy\) MUST be a multiple of which of the following, not COULD be a multiple.
\(x\) is a multiple of 4 --> \(x=4m\), for some positive multiple \(m\), so \(x\) can be: 4, 8, 12, ...
\(y\) is a multiple of 6 --> \(y=6n\), for some positive multiple \(n\), so \(y\) can be: 6, 12, 18, ...
So, \(xy=(4m)*(6n)=24mn\), so \(xy\) is in any case a multiple of 24, which means it must be a multiple of 8 and 12, but not necessarily of 18. For example the least value of \(x\) is 4 and the least value of \(y\) is 6, so
the least value of \(xy\) is 24, which is a multiple of both 8 and 12, but not 18.
Answer: B.
Hope it's clear.
-- i wrote down, x = 4 (given x is a multiple of 4) and y = 6 (given y is a multiple of 6)
-- For x = 4 and y =6 ; In this case, the Lowest common multiple between these two numbers is actually 12 { I set to calculate the L.C.M as i thought this will give me a multiple of those two numbers }
-- Thus i thought, a L.C.M of 12 for x =4 and L = 6 ....thus the multiple must be 12 | 24 | 36 | 48 | 60 | 72 ......