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Remainder y-3

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Re: Remainder y-3

by logitech » Tue Nov 18, 2008 11:23 pm
Abdulla wrote:when 15 divided by y, the remainder is y-3. if y must be an integer, what are all possible values of y ?
15 = a Y + (Y-3)

18 = Y ( a + 1 )


Possible Y values : ( -18, -9, -6, -3, -2, -1, 1, 2, 3, 6, 9, 18 )
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by cramya » Wed Nov 19, 2008 6:41 am
Logitech,
I am not sure on some of the values of y indicated above

I can understand the 9 and 6.

Eg: 18

Can u explain further how u came ip with some of the y values above.

15/18 quotient is 0 remainder is 18 which is not equal to y-3 = 18-3 = 15

Am I missing something??
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by jimmiejaz » Wed Nov 19, 2008 9:29 am
cramya wrote:Logitech,
I am not sure on some of the values of y indicated above

I can understand the 9 and 6.

Eg: 18

Can u explain further how u came ip with some of the y values above.

15/18 quotient is 0 remainder is 18 which is not equal to y-3 = 18-3 = 15

Am I missing something??
Cramya,
The remainder will be 15.eg. 2/5 remainder is 2 and not 3. :)
Hope this helps!!!!
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by jimmiejaz » Wed Nov 19, 2008 9:31 am
i guess you have solved too many problems.....
:)
Anyways u guys rock!!!!!!!!!
keep posting ......
This time i am sure all of us :) will clear the GMAT.
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by cramya » Wed Nov 19, 2008 9:51 am
I guess you have solved too many problems.....
I guess I should take this as the omen to move to Verbal and focus less on Quant....

Thanks Jimmie!

:-)
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Re: Remainder y-3

by uttara » Wed Nov 19, 2008 10:48 am
logitech wrote:
Abdulla wrote:when 15 divided by y, the remainder is y-3. if y must be an integer, what are all possible values of y ?
15 = a Y + (Y-3)

18 = Y ( a + 1 )


Possible Y values : ( -18, -9, -6, -3, -2, -1, 1, 2, 3, 6, 9, 18 )

in 18 = Y ( a + 1 )

a can't be a negative number since Quotient can't be negative
&
y-3>=0 since Remainder can't be negative

Possible Y values : (18,9,6,3)
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by logitech » Wed Nov 19, 2008 2:51 pm
Thanks Uttara. I will keep that in mind. Are we all clear on this question ?
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by Abdulla » Wed Nov 19, 2008 4:02 pm
Logitech your explanation make sense to me, but why Uttara is giving another answers .. and you said you will keep that in mind ... who is right?
Abdulla
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by logitech » Wed Nov 19, 2008 4:09 pm
Abdulla wrote:Logitech your explanation make sense to me, but why Uttara is giving another answers .. and you said you will keep that in mind ... who is right?
Uttara is correct.

The remainder needs to be greater than ZERO. So

y-3 > 0

y > 3
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Re: Remainder y-3

by 4meonly » Sun Nov 23, 2008 7:18 am
uttara wrote: Possible Y values : (18,9,6,3)
Possible Y values : (18,9,6,3, 2 and 1)
if a=8 and 17 corresp.
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Re: Remainder y-3

by logitech » Sun Nov 23, 2008 8:39 am
4meonly wrote:
uttara wrote: Possible Y values : (18,9,6,3)
Possible Y values : (18,9,6,3, 2 and 1)
if a=8 and 17 corresp.
I thought that you supposed to be greater than three in this question ?
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by 4meonly » Mon Nov 24, 2008 2:00 am
y-3>=0 since Remainder can't be negative
Yes, thanx.
My mistake.
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