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remainder - very interesting question

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remainder - very interesting question

by SofesMary » Wed Mar 04, 2009 9:38 am
Hi all,

Can you please help me on this one?

Thanks!!!

When positive integer n is divided by 3, the remainder is 2; when positive integer t is divided by 5, the remainder is 2. What is the remainder when the product is nt is divided by 15?

a) n-2 is divisible by 3

b) t is divisible by 3
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by DanaJ » Wed Mar 04, 2009 10:13 am
I initially wanted to solve this problem algebraically, but I soon realized it wouldn't take me far.
So I started plugging in numbers...
Now, let's analyze each stmt:

1. provides no new info: it's obvious that, since the remainder of dividing n to 3 is 2, n - 2 will be divisible by 3. This means that 1 is insufficient.

2 was a bit more challenging, but I ended up plugging in the smallest numbers I could think of. This is why I ended up with two cases that were both consistent with the restrictions of the problem, but provided different answers:
a. n = 5
t = 27
nt = 27 * 5 = 3 * 9 * 5 = 15 * 9 - in this case, nt is divisible by 15, so the remainder will be 0.

b. n = 14
t = 27
nt = 27 * 14 - in this case, nt is obviously not divisible by 15 (since we don't have a 5 in there somewhere), so the remainder is not going to be equal to 0, like in the previous case. We're not even interested into finding out this remainder, since knowing that it is different from 0 is enough to tell that there are two possible cases for stmt 2.

Put both stmts together and you get nothing still, since we've already established that stmt 1 has no value.

So my guess will be E

What is the OA?
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by marcusking » Wed Mar 04, 2009 10:57 am
DanaJ wrote:I initially wanted to solve this problem algebraically, but I soon realized it wouldn't take me far.
So I started plugging in numbers...
Now, let's analyze each stmt:

1. provides no new info: it's obvious that, since the remainder of dividing n to 3 is 2, n - 2 will be divisible by 3. This means that 1 is insufficient.

2 was a bit more challenging, but I ended up plugging in the smallest numbers I could think of. This is why I ended up with two cases that were both consistent with the restrictions of the problem, but provided different answers:
a. n = 5
t = 27
nt = 27 * 5 = 3 * 9 * 5 = 15 * 9 - in this case, nt is divisible by 15, so the remainder will be 0.

b. n = 14
t = 27
nt = 27 * 14 - in this case, nt is obviously not divisible by 15 (since we don't have a 5 in there somewhere), so the remainder is not going to be equal to 0, like in the previous case. We're not even interested into finding out this remainder, since knowing that it is different from 0 is enough to tell that there are two possible cases for stmt 2.

Put both stmts together and you get nothing still, since we've already established that stmt 1 has no value.

So my guess will be E

What is the OA?
Did you mean to make t=21?
27/3 is 9 -> 9/5 doesn't give a remainder of 2.

Instead if you make t=21
nt = 21*5 = 105 / 15 = 7 Remainder 0

Other than that your logic makes sense
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by DanaJ » Wed Mar 04, 2009 11:01 am
t = 21 is not consistent with "when positive integer t is divided by 5, the remainder is 2". The remainder when dividing t = 21 by 5 will be 1.
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by marcusking » Wed Mar 04, 2009 11:02 am
DanaJ wrote:t = 21 is not consistent with "when positive integer t is divided by 5, the remainder is 2". The remainder when dividing t = 21 by 5 will be 1.
Sorry I got t and n mixed up. You are correct :oops:
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by GMATaGoGo » Wed Mar 04, 2009 12:18 pm
To solve this algebraically:

Using initial statements in the question stem, you get
n=5q+2
t=5K+2
where q and k are constants

nt/15= (5q+2)(5k+2)/3.5
that works out to be (25qk+10q+10k+4)/3.5
Look closely, and you'll see that all you need to know is what is k.

1. n-2 is divisible by 3
=5q+2. useless so, not suff

2. t is divisible by 3

saine t=5k+2, this means K cannot be a multiple of 3. K can be either 2 or 5. not suff

Combine the 2, you still dont know what K is. So E.
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by DanaJ » Wed Mar 04, 2009 1:01 pm
GMATaGoGo wrote:To solve this algebraically:

Using initial statements in the question stem, you get
n=5q+2
t=5K+2
where q and k are constants
The problem here is that n = 3q + 2.
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by pakaskwa » Wed Mar 04, 2009 8:57 pm
I tried this way:
n=3p+2
t=5q+2

1st condition doesn't mean anything. It simply repeated what is already available in the question. No matter how you proceed, you'll have to use 1st condition.

Now we focus on 2nd condition. If t is divisible by 3, we can plug in first number 12 (12=5x2+2). And we got:
nt = 12(3q+2)

Now we can easily see that when we change the value of q, nt is not necessarily divisible by 15. So answer is E.
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by mrsmarthi » Wed Mar 04, 2009 9:35 pm
This is how I solved the question.

Following information is given in the question stem.

a) n is divisible by 3 leaving a remainder of 2 ==> (n-2) is divisible by 3.
b) t is divisible by 5 leaving a remainder of 2 ==> (t-2) is divisible by 5.

Question being asked is what is the remainder when nt is divided by 15.

From stmt 1: n-2 is divisible by 3. This is redundant information since it is already provided in the Question stem. Hence insufficient.

From Stmt2: t is divisible by 3. Also we can conclude that t is not divisible by since it is tiven t-2 is divisible by 5. So the remainder when nt when divided by 15 depends on whether 5 is a factor of n or not. But this statment does not provide this information. Hence insufficient.

Combining both the statments also does not provide if n is divisible 5 or not.

IMO E.
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by chicham » Mon Mar 23, 2009 1:27 am
Hello guys,
I just had this question,
and the official answer is C not E,
I had answered "E", I don't get it//
Weird!

Anyone?
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by 2ndShot » Mon Mar 23, 2009 9:02 am
Are you sure this is not a printing error?
The only thing I can infer from the question is that the remainder has to be a multiple of 3, but that does not answer the question.
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by Tryingmybest » Mon Mar 23, 2009 9:26 am
Similar official GMAT Problem was discussed in this link months back. Please check

https://www.beatthegmat.com/remainder-t23606.html
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