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If r is the remainder when the positive integer n is divided

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If r is the remainder when the positive integer n is divided [#permalink] New post 02 Jul 2008, 03:42
If r is the remainder when the positive integer n is divided by 7, what is the value of r?
(1) When r is divided by 21, the remainder is an odd number.
(2) When n is divided by 28, the remainder is 3.

n = 7x + r

according to (1):
n = 21y + r1 (r1 is odd)
n = 7(3y) + r1
meaning the remainder while dividing by 21 and by 7 is the same odd number, but we don't have a way to find it.

according to (2):
n = 28y + 3
n = 7(4y) + 3
therefore r = 3, and (2) is sufficient.

Is my reasoning correct, is there a better way to explain it? Of course, we don't to explain in the GMAT, but I want to make sure my understanding is correct.
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Re: remainder question - is my reasoning correct [#permalink] New post 02 Jul 2008, 03:50
nirimblf wrote:
according to (1):
n = 21y + r1 (r1 is odd)
n = 7(3y) + r1
meaning the remainder while dividing by 21 and by 7 is the same odd number, but we don't have a way to find it.

Nope, r and r1 are not necessarily the same. Don't forget :
n = 7x + r, with r between 0 and 6 (inclusive)

n = 21y + r1, with r1 between 0 and 20 (inclusive)

We could have r1=17, which is odd and though r won't be equal to 17 then (but 17-2*7 = 3)

That being said, it its true that you don't know much about r1 and therefore don't know much about r (all you can say is that it is odd too, so r belongs to {1,3,5})

For instance :
n = 26 ==> r1 = 5 and r=5
n = 38 ==> r1 = 17 and r=3

==> (1) is insufficient

Last edited by Oski on 02 Jul 2008, 03:52, edited 1 time in total.
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Re: remainder question - is my reasoning correct [#permalink] New post 02 Jul 2008, 03:51
nirimblf wrote:
If r is the remainder when the positive integer n is divided by 7, what is the value of r?
(1) When r is divided by 21, the remainder is an odd number.
(2) When n is divided by 28, the remainder is 3.

n = 7x + r

according to (1):
n = 21y + r1 (r1 is odd)
n = 7(3y) + r1
meaning the remainder while dividing by 21 and by 7 is the same odd number, but we don't have a way to find it.


well you really dont need to get into calculations here, we know r < 7 so the reminder of r/21 is "r" for sure and r being odd r can be 1,3,5 (NOT sufficient)

Quote:
according to (2):
n = 28y + 3
n = 7(4y) + 3
therefore r = 3, and (2) is sufficient.

good approach i think..

Quote:
Is my reasoning correct, is there a better way to explain it? Of course, we don't to explain in the GMAT, but I want to make sure my understanding is correct.

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Re: remainder question - is my reasoning correct [#permalink] New post 02 Jul 2008, 03:59
Oski wrote:
nirimblf wrote:
according to (1):
n = 21y + r1 (r1 is odd)
n = 7(3y) + r1
meaning the remainder while dividing by 21 and by 7 is the same odd number, but we don't have a way to find it.

Nope, r and r1 are not necessarily the same. Don't forget :
n = 7x + r, with r between 0 and 6 (inclusive)

n = 21y + r1, with r1 between 0 and 20 (inclusive)

We could have r1=17, which is odd and though r won't be equal to 17 then (but 17-2*7 = 3)

That being said, it its true that you don't know much about r1 and therefore don't know much about r (all you can say is that it is odd too, so r belongs to {1,3,5})

For instance :
n = 26 ==> r1 = 5 and r=5
n = 38 ==> r1 = 17 and r=3

==> (1) is insufficient


So the reason we can move from
n = 28y + 3
to
n = 7(4y) + 3
and deduce the remainder is the same, is that 3 < 7 ?
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Re: remainder question - is my reasoning correct [#permalink] New post 02 Jul 2008, 04:14
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nirimblf wrote:
So the reason we can move from
n = 28y + 3
to
n = 7(4y) + 3
and deduce the remainder is the same, is that 3 < 7 ?

Exactly :wink:
Re: remainder question - is my reasoning correct   [#permalink] 02 Jul 2008, 04:14
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