If m and n are nonzero integers, is m/n an integer?

[brian.lange]

Here are two problems that I believe test the same concept. I am having difficulty as I will explain

If m and n are nonzero integers, is m/n an integer?
(1) 2m is divisible by n
(2) m is divisible by 2n

I came across a similar problem in the OG, that shows why I am having difficulty with this concept

[Note: portion deleted by moderator; please do not quote from the OG here!]

Now, in the explanation for the 1st problem, statement 1 is ruled insufficient. Although the elements of n are in 2m's prime box, m alone may not have all of the elements of n's prime box. Thus n may or may not be a factor of m.

But in the second problem, I have essentially the same concept, but [deleted - see note above]

So to summarize
- problem 1 - 2m/n is integer, and m may or may not be divisible by n
- problem 2 - [deleted]

Aren't these two problems saying two different things? What am I missing?

[JonathanSchneider]

I had to delete the portion of your post that quoted from the OG. We cannot post from the OG here. However, I've still tried to answer your question:

The difference is that in the first problem, the 2 is in the numerator, whereas in the second problem, the 30 is in the denominator. This makes all the difference.

Another way to see it:

Statement 1 from Problem 1:

2m/n = i

For these divisibility problems, simply placing an "i" can be illuminating. Here "i" just represents "some integer." We are looking to see whether m/n = an integer. So, manipulate to get m/n by dividing by 2, and you will see that m/n = i/2. Because not all integers, when divided by 2, yield another integer, the answer is insufficient.

[brian.lange]

That is somewhat illuminating...for the CAT problem...but still leaves my wondering about the OG problem.

Here is my own example:

If 5n/28 = an integer, why must n be divisible by 28?

If I look at it in the manner you described above, I would say 5n/28 = k, where k is an integer. If I divide both sides by 5, I get n/28 = k/5. So I would say that n/28 = an integer / by 5, which doesn't prove to me that n is divisible by 28.

How can I look at this problem in the same way you explained the CAT problem?

[cyapt81]

so what's the answer to OP's 1st problem? Is it B?

[esledge]

so what's the answer to OP's 1st problem? Is it B?
so what's the answer to OP's 1st problem? Is it B?

Yes, the correct answer is B.

Here is my own example:

If 5n/28 = an integer, why must n be divisible by 28?

If I look at it in the manner you described above, I would say 5n/28 = k, where k is an integer. If I divide both sides by 5, I get n/28 = k/5. So I would say that n/28 = an integer / by 5, which doesn't prove to me that n is divisible by 28.

How can I look at this problem in the same way you explained the CAT problem?

You can look at this problem in a similar way as Jonathan did the CAT problem. The difference is that you have two actual numbers (5 and 28), so whether those numbers have shared factors will be relevant, as we will see.

IMPORTANT: Your example would also have to specify that n itself is an integer!

Let's manipulate to actually isolate n:
5n/28 = k, where k is some integer.
n = 28k/5, where k is some integer.

Thus, n could be 0, 28/5, (28*2)/5, (28*3)/5, (28*4)/5, 28, (28*6)/5, etc.

n will only be an integer when k = a multiple of 5, because 28 and 5 have no shared factors. The only way to cancel the 5 in the denominator is for the "some integer" to do so! Going back to our equation with this observation...

n = 28k/5, where k is some multiple of 5.
n = 28 * (mult of 5)/5
n = 28 * j, where j is some integer.
n is a multiple of 28.