GPrep - If n is a positive integer and r is the remainder

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by ErikaPrepScholar » Thu Sep 28, 2017 5:47 am
The word "remainder" tells us that we're dealing with, what else, a remainder problem. Remainder problems scare a lot of students because they don't involve an easy to use/memorize formula. However, this means that we have a great opportunity to plug in numbers.

Even though this isn't technically a "word problem", we still need to translate the words into math to build an equation:
(n-1)(n+1)/24 = ? | R: r
Let's make a note that n must be a positive integer and move on to our statements.

Statement I
This statement tells us that n is not divisible by two - in other words, it's telling us that n is odd. Let's try plugging in numbers. When we select numbers to plug in, our goal is to prove that the statement is insufficient: in other words, we want to pick numbers that will give us different results. We also want to pick numbers that are easy to work with to save time.

We see that one of the values in our numerator is (n-1), which means that picking 1 will give us a zero in our numerator. That seems like it'll give us an interesting result, so we'll give it a shot:

(1-1)(1+1)/24
(0)(2)/24
0/24
0 | R: 0

So when n=1, r=0. Let's try our next odd number up, 3 - based on the size of the denominator, it seems like our numerator will be smaller than the denominator, giving a solution of 0 with positive remainder:

(3-1)(3+1)/24
(2)(4)/24
8/24
0 | R: 8

So when n=3, r=8. This means that r can be either 0 or 8 given Statement 1. Since we can't find a single value for r, Statement 1 is insufficient.

Statement II
This statement tells us that n is not divisible by three. That knocks n=3 out of the running. n=1 still works, however, so we know that r=0 is still a possibility given Statement 2.

Since we tried only odd numbers last time, let's try an even number this time to see if that changes things up: we'll do 2 to keep our numbers easy to work with:

(2-1)(2+1)/24
(1)(3)/24
3/24
0 | R: 3

So when n=2, r=3. This means that r can be either 0 or 3 given Statement 2. Like before, since we can't find a single value for r, Statement 2 is insufficient.

BOTH
Putting these two statements together, we know that n must be odd and cannot be divisible by 3: so we have 1, 5, 7, 11, etc. These numbers are going to get pretty big pretty fast, so let's try them from smallest to greatest. We already know that r=0 when n=1, so we want to find a positive value for r to prove that both statements are insufficient:

(5-1)(5+1)/24
(4)(6)/24
24/24
1 | R: 0

So when n=5, r=0. That's the same as when n=1. Let's try the next number up, 7:

(7-1)(7+1)/24
(6)(8)/24
48/24
2 | R: 0

So when n=7, r=0. We're starting to see the hints of a pattern here. Let's try one more, 11, to be sure:

(11-1)(11+1)/24
(10)(12)/24
120/24
5 | R: 0

So when n=7, r=0. Once we've tried at least 4 numbers in a series and confirmed that we've done a reasonable job picking numbers that would give us different results, we can usually determine that we have a pattern. Here, we can say confidently that given Statement 1 and Statement 2, r will always be 0. This means that the correct answer is C: BOTH statements together are sufficient.

We actually featured this problem recently on the PrepScholar GMAT blog as one of the 5 Hardest Data Sufficiency Questions. I recommend checking out the article for more strategies and trends we can take away from this and other 700+ level problems!
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