Factor Question

I got E. However, it took me several minutes to solve, so there has to be a better way.


Where are you finding these problems?

If K is a multiple of 11, is KL a multiple of 154 ?

K=11*m with m a constant
KL=11*m*L

1) L has all same factors as 42
42=2*3*7
So L has at least these factors: 2, 3, 7
In that case L only has these factors, we can write:
KL=11*m*2*3*7
KL=154*3*m because 154=11*7*2 (you can start by looking at what are the factors of 154 in order to know what you are looking for)
So you see that KL is a multiple of 154.
Sufficient.

2) K is divisible by 21
It also means K is a multiple of 21, which can be written like that:
K=21*p (with p a constant)
K=11*m with m a constant, that we have from the beginning
You can see that K is a multiple of 21 and 11
So K will have among its factors: 11, 7, 3
(154=11*7*2, so we need a 2) The problem is that we do not know if K has a factor of 2, L could also have or not a 2. So we do not know if KL is a multiple of 154.
[For example imagine the smallest value of K which is 21*11=231. That is a not a multiple of 154. If L=2 so KL = 462 and it is a multplie of 154. If L=3, KL is not a multiple of 154]
Insufficient.

So it is A for me.
That's not so long, but I can be wrong. The important thing is to translate "x is a multiple of y" into "x=y*k with k a constant".
Tell me if it's ok.

hi Pepp.
Thanks for replying to question. Question appeared in PR manual ( which they give to u if u attend their course)

Well i brainstormed and find that since L has same same factor as of 42 so L would be "42". None of other number would share the same number of factors.

Factors of 42 are 1 ,2 ,3 ,6,7, 14 , 21, 42

So if K = 11 (1)
L= 42

Quetsion is that would KL = 11(1) * 42 be mutiple of 154 . And we can deduce that 11(1) * 7*6 is divisible by 154 and thus makes a mutile pf 154.

Option A is clear.

By #2 clue , we wont get value of L so Crosse option D.

Ansawer should be A.

"All same factors":
same in type of factors?
or same in type and number of factors?

In the question, it does not change anything but I would be glad to know.

By factors of 42, I take that literally: 1, 2, 3, 6, 7, 14, 21, 42. At least these factors. I (also) first ran across this type of problem in a Princeton Review book, so credit to those guys.

Question, simply: If KL has the factors of 154, it must be a multiple of 154. They've given us 11. All we need now is 7 and 2.

(1) gives us 7 and 2. Sufficient.
(2) gives us only 7 (and 3, which we don't need). Insufficient.

I believe that same factors means that number of factors would be same and those factors would be distinctive in nature.

Same factors as 42 should imply that no itself is 42. Otherwise we would keep on getting several no having same no of factrors as 42.