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## Remainders Data Sufficiency Problem

tagged by: mbaorbust3

This topic has 1 member reply
mbaorbust3 Newbie | Next Rank: 10 Posts
Joined
25 Aug 2014
Posted:
4 messages

#### Remainders Data Sufficiency Problem

Wed Dec 06, 2017 12:40 pm
have a question about a problem found in Chapter 5 of the GMAT Advanced Quant Strategy Guide, Second Edition. Problem 7 (page 161) reads:

If X is an integer, what is the remainder when X is divided by 5?

1) X^2 has a remainder of 4 when divided by 5
2) X^3 has a remainder of 2 when divided by 5

The OA answer is B with X=3. However, X= 3 and 8 for both statements produce remainders of 4 and 2 for X^2 and X^3 respectively. So shouldnâ€™t the correct answer be E? Please advise. Thanks.

OWN Junior | Next Rank: 30 Posts
Joined
10 Nov 2017
Posted:
13 messages
7
Wed Dec 06, 2017 4:10 pm
Hi mbaorbust3,

I think you might have missed what the question was actually asking. The questions asks what the REMAINDER of X/5 is. NOT what X is (as such, X = 3 and X = 8 both give us the same answer since they both give a remainder of 3 when divided by 5).

How to solve this question using a nifty rule (I provided the proof for this rule below) and some testing:
Background:
X is some integer and has a remainder when divided by 5 (this remainder can either be 0,1,2,3, or 4 as far as we know)
As such X can be re-written as X = 5N + R, where N can be any integer (including 0) and R is either 0 , 1, 2, 3, or 4.
You can see that dividing X by 5 gives us (N + R/5) or NrR

Fact 1: the remainder of X^2 /5 is 4
Right off the bat, I know that the remainder is not 0, so I can get rid of that as an option
I know that if X has a remainder of R when divided by 5, the remainder of X^2/5 will be equal to the REMAINDER of R^2/5.
So let's do some quick math for all of our options for R (1,2,3,4)
If R = 1, R^2 = 1, 1/5 gives 0r1, so the remainder isn't 4
If R = 2, R^2 = 4, 4/5 gives 0r4, the remainder IS 4, so this could be a match
If R = 3, R^2 = 9, 9/5 gives 1r4, again the remainder is 4, so this could also be a match and we can stop here
Basically, with the information we have we can't say whether the remainder when X is divided by 5 is 2 or 3, so fact 1 is insufficient.

Fact 2: the remainder of X^3/5 is 2
Here, we can also rule out zero as the remainder
Using the same theory, let's test out values of R:
If R = 1, R^3 = 1, 1/5 gives 0r1, so the remainder isn't 2
If R = 2, R^3 = 8, 8/5 gives 1r3, so the remainder isn't 2
If R = 3, R^3 = 27, 27/5 gives 5r2, the remainder IS 4, so this could be a match
If R = 4, R^3 = 64, 64/5 gives 12r4, so the remainder isn't 2
In this case, only R=3 is a match so we know that R =3. Which means that X could equal {3,8,13,18,......} as far as we know.

So, we get Answer Choice B.

Here is the full algebraic approach for this question. It's nasty and I personally would not use it on the test.

Background:
X is some integer and has a remainder when divided by 5 (this remainder can either be 0,1,2,3, or 4 as far as we know)
As such X can be re-written as X = 5N + R, where N can be any integer (including 0) and R is either 0 , 1, 2, 3, or 4.
You can see that dividing X by 5 gives us (N + R/5) or NrR

Fact 1: the remainder of X^2 /5 is 4
Based on this fact, we can rule out 0 as the remainder
Let's express this as follows
$$\frac{\left(5N+R\right)^2}{5}\ \ =\ \frac{25N^2+10NR\ +R^2}{5}$$
It isn't difficult to see that in this case, 25N^2 and 10NR will both be divisible by 5, so let's look at R^2.
We know that R^2/5 gives us a remainder of 4, from the prompt.
We also know that R could be 0, 1, 2, 3, or 4 (from the question). Let's square these
If R = 1, R^2 = 1, 1/5 gives 0r1, so the remainder isn't 4
If R = 2, R^2 = 4, 4/5 gives 0r4, the remainder IS 4, so this could be a match
If R = 3, R^2 = 9, 9/5 gives 1r4, again the remainder is 4, so this could also be a match and we can stop here
Basically, with the information we have we can't say whether the remainder when X is divided by 5 is 2 or 3, so fact 1 is insufficient.

Fact 2: the remainder of X^3/5 is 2
Here, we can also rule out zero as the remainder
So basically when we divide (5N+R)^3 by 5, we get 2
I'm going to take a shortcut here from the previous fact and say that when I cube that expression, the coefficient of N^3, N^2, and N will be divisible by 5.
Therefore I know the R^3/5 gives me a remainder of 2. So now I try all the different R:
If R = 1, R^3 = 1, 1/5 gives 0r1, so the remainder isn't 2
If R = 2, R^3 = 8, 8/5 gives 1r3, so the remainder isn't 2
If R = 3, R^3 = 27, 27/5 gives 5r2, the remainder IS 4, so this could be a match
If R = 4, R^3 = 64, 64/5 gives 12r4, so the remainder isn't 2
In this case, only R=3 is a match so we know that R =3. Which means that X could equal {3,8,13,18,......} as far as we know.

Disclaimer: I would not do this on the exam unless this was at the very end and I had some extra time to do all this algebra.

Not only is there an algebraic/rule-based approach to this question, but also an empirical one: You can simply test a bunch of values of X by squaring/cubing them and if you get 3 or 4 consistent answers for a fact, then there is a good chance it would be sufficient. Not as airtight as knowing how the pattern works, but great in a pinch.

Hope this helped!

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