If m & n are integers and m > n, is m divisible by n?

[mclaren7]

Dear moderators and friends,

Data sufficiency:

If m & n are integers and m > n, is m divisible by n?
1. 2m is divisible by n.
2. m^2 is divisible by n.

For 1,
2m / n --> assuming n is 2, m could by 3, 4.
insuff.

For 2,
m.m / n
I do not know if there is a mathematical way of solving, I just plugged in numbers.
Works for 4.4/2 ---> 4/2
Not for 6.6/4 ---> 6/4
insuff

Together, i am not sure how best to proceed.
m^2 - 2m / n
= m(m-2)/n

Thus m or (m-2) is divisible by n.
???

Source: paid GMAT services. Is this kind of question too hard?

Thanks
KH

[Guest]

I think this type of question is quite common.

my answer is E.

I plugged in numbers too, like you did.

For questions like this, i like to make a little table.

1.)
m 2m n
6 12 4 Doesnt work
8 16 4 Works

So, insufficient.
2.)
m m^2 n
6 36 4 Doesnt work
8 64 4 Works

Now, together
m m^2 2m n
6 36 12 4 Doesnt work
8 64 16 4 Works

So, answer is E.

[mclaren7]

Much appreciated.
Thanks & good luck.

[RonPurewal]

since this question is about divisibility by specific numbers (even though those 'numbers' are variables, in this case), you should employ the 'prime box' (prime factorization) approach.

QUESTION:
does the 'prime box' {prime factorization) of m contain all the primes that multiply to yield n?

(1)
this means that the 'prime box' of 2m contains all the primes that multiply together to make n. (IMPORTANT: n itself doesn't have to be in the box, as it is not necessarily prime.)
the 'prime box' of m is the same as the above 'prime box', except for that one '2' has been removed.
if that '2' is part of n, though - and there aren't any other '2's to take its place - then all the primes needed to make n will no longer be present (answer to question = NO).
if n does not contain a 2, though, or if there are other '2's to take the place of the removed one, then all the primes required to make n will still be in the box for m (answer to question = YES).
INSUFFICIENT

(2)
this means that the 'prime box' of m^2 contains all the primes that multiply together to make n. (as before, n itself doesn't have to be in the box, as it doesn't have to be prime.)
the prime box for m^2 contains all the primes required to make m, TWICE. if n is contained in this box, we have no guarantee that it will still be contained in the smaller box for just m (which contains only half as many primes).
INSUFFICIENT

together
taking both statements together doesn't resolve the issues stated for either.
still INSUFFICIENT

answer = e

--

follow-up: if you're told that n is prime, then the answer is now B. see if you can explain that and/or come up with examples to substantiate it.