MGMAT Math question

This topic has expert replies
Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Sun Sep 16, 2007 6:43 am

MGMAT Math question

by PL » Mon Oct 01, 2007 10:44 pm
Hi

I took a MGMAT CAT yesterday and I am having trouble with the explanation they provide for below question.

Is n/18 an integer?

(1) 5n/18 is an integer.
(2) 3n/18 is an integer.

I think the right answer for this one is A (statement 1 is sufficient), but MGMAT explanation says the correct answer is C (Both statements TOGETHER are sufficient).

Can someone explain.

Regards,

- PL

Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Tue Sep 25, 2007 11:35 am

by homer » Tue Oct 02, 2007 9:29 am
The trick (or catch) is that n itself may not be an integer.

So, if n = (18/5) condition 1 is true - but n/18 = 1/5 => not an integer

The second condition by itself is also not sufficient - as yuo have probably figured out anyways. BUT it tels you that n/6 is an integer. This establishes that _n_ is an integer.

Now - if we go back to condition 1, with the knowledge, that n is an integer, n/18 is an integer is tautologically true.

Hence, C - both conditions are necessary.

Master | Next Rank: 500 Posts
Posts: 321
Joined: Tue Aug 28, 2007 5:42 am
Thanked: 1 times

by kajcha » Tue Oct 02, 2007 9:33 am
PL,

Stmt 1 - 5n/18 is not suff.

suppose n/18 can be 1 or it can be 0.6. In both case 5n/18 is an integer.

Stmt 2 - n/18 can be 1 or can be 0.2. Again, in both cases 3n/18 is an integer.

together, SUFF as n/18 has to be integer for 5n/18 and 3n/18 to be integer.

Ans C

Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Sun Sep 16, 2007 6:43 am

by PL » Tue Oct 02, 2007 9:39 am
homer wrote:The trick (or catch) is that n itself may not be an integer.

So, if n = (18/5) condition 1 is true - but n/18 = 1/5 => not an integer

The second condition by itself is also not sufficient - as yuo have probably figured out anyways. BUT it tels you that n/6 is an integer. This establishes that _n_ is an integer.

Now - if we go back to condition 1, with the knowledge, that n is an integer, n/18 is an integer is tautologically true.

Hence, C - both conditions are necessary.

Good explanation. Thanks

Master | Next Rank: 500 Posts
Posts: 184
Joined: Sat Apr 14, 2007 9:23 am
Location: Madison, WI
Thanked: 17 times

Re: MGMAT "factor box" method

by ldoolitt » Wed Oct 10, 2007 4:19 pm
rosenjon wrote:I found MGMAT's factor box explanation very helpful for these types of problems. This is a slightly less "math intensive" way to approach it.

Remember that any number is a factor of another number if that number has the same number and type of primes. MGMAT has you put these primes in a "factor box" that represents that number. This makes it very easy to break down the problem and compare factor boxes to see if a number is a factor of another.

So let's look at the question:

Is n/18 an integer? Well, before we even get to the question, let's factor 18.

18 = 9x2 = 3x3x2

So let's rethink the question stem? Is n/(3x3x2) an integer. Or, in other words, if n has 3,3,2 in its "factor box", then n is an integer. If n doesn't have 3,3,2 in its factor box, then n/18 cannot be an integer.

Statement 1 says that 5n/18 is an integer. Well, we know that the prime factorization of 18 = 3x3x2 from above. Since 5n IS AN INTEGER, then 5n must have 3,3,2 in its "factor box". However, since 3,3,2 MUST be a factor of 5n, and 5 is not divisble by 3 or 2, then n must be divisible by 3,3,2.

Our stem statement says that n is divisible by 18 if it has 3,3,2 in its factor box. Statement 1 proves that it does, so statement 1 is sufficient.

Statement 2 asks if 3n/18 is an integer.

The explanation for statement 2 is along the same lines as above.

3n/18 IS an integer.

Therefore, 3n/(3x3x2) must be an integer. If you reduce this, you get n is divisible by 3x2. Therefore, according to this statement, n has 3,2 in its factor box.

However, before we began the problem, we decided that n must have 3,3,2 in its factor box to be divisible by 18. Statement 2 only tells us that it has 3,2.

Therefore, n MIGHT be divisible by 18. We don't know the value of n, so we can't determine whether it is or not. If we knew that n ONLY had 3,2 in its factor box, and nothing else, then the answer would be no, and 2 would be sufficient.

But n has some of the factors of 18 in its factor box, plus some other set of unknown factors. Therefore, n could be divisible by 18, or it might not be. Without knowing the specific value of n, it is impossible to determine.

Therefore, Statement 2 is not sufficient.

The answer is therefore A.
That is a phenominal explanation. That factor box concept helps out a lot.

Senior | Next Rank: 100 Posts
Posts: 50
Joined: Thu Oct 04, 2007 12:26 pm
Location: RTP, NC
Thanked: 26 times
Followed by:3 members

by achandwa » Mon Oct 22, 2007 1:14 pm
Excellent explaination rosenjon!!! Thanks.

Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Thu Oct 18, 2007 8:05 am

Answer is C?

by gsaxton » Wed Oct 31, 2007 8:15 am
I too really like rosenjohn's "factor box" explanation, but in this case must agree with the first reply:

Statement 1) only tells us that 5n has 3,3,2 in its "factor box," but it could have this by n being, for example, 18/5. In that case, (5*18/5)/18 is an integer, but n/18 is not (it would be 1/5).

In short, statement 1 is insufficient and, as far as I can see it, the answer is C.

Senior | Next Rank: 100 Posts
Posts: 50
Joined: Thu Oct 04, 2007 12:26 pm
Location: RTP, NC
Thanked: 26 times
Followed by:3 members

by achandwa » Wed Nov 21, 2007 8:55 pm
If n is an integer, rosenjon's explaination would hold. The answer would be A.

But since n is not given to be an integer, per Gsaxton's reasoning, (1) alone is not sufficient. (2) alone is not sufficient along the same lines.
But (1) and (2) taken together, are sufficient. Because if both 5n/18 and 3n/18 are integers, then n has to be a multiple of 18. Try any other possibility and one of the two or both of 3n/15 and 5n/18 will not turn out to be an integer. Thus, the answer is C.