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DS-Remainder of Quad Equation

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DS-Remainder of Quad Equation [#permalink] New post 27 Sep 2008, 23:49
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Re: DS-Remainder of Quad Equation [#permalink] New post 29 Sep 2008, 00:21
A

if you just post the picture instead of a word doc, you would get more answers, not to mention, it would be easier too.
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Re: DS-Remainder of Quad Equation [#permalink] New post 30 Sep 2008, 01:34
to our experts - please help us with this out!!
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Re: DS-Remainder of Quad Equation [#permalink] New post 30 Sep 2008, 01:47
From stmt1, t = 7n + 6 for n = 0,1,2,.....

Hence, t^2 + 5t + 6 = 49n^2 + 119n + 72
When you divide this expression by 7, the remainder will be 2. Hence, sufficient.

For stmt2, t^2 = 7n + 1 for n = 0,1,2,3....
Hence, t = sqrt(7n+1) and since t is an integer, hence, it can only be 1,5,9,....

For t = 1, the given expression will have remainder as 5.
For t = 5, the given expression will have remainder as 0.

Hence ,stmt2 is not sufficient.
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Re: DS-Remainder of Quad Equation [#permalink] New post 02 Oct 2008, 09:10
vr4indian wrote:
can anyone explain this?

IMO A
1) t divided by 7 leaves remainder 6.
t^2 divided by 7 leaves remainder same as 36/7 & i.e. 1
5t divided by 7 leaves remainder as 5*6/7 i.e. 2
6 divided by 7 leaves remainder as 6.
All 3 plus leaves remainder same as (1+2+6)/7 or 9/7 & i.e. 2

2) With this we cant calculate remainder for 5t. Hence insufficient.

Logic used: If X divided by A leaves remainder as x & Y divided by A leaves remainder as y, then X+Y when divided by A leaves the same remainder as (x+y)/7 would.
Also, XY when divided by A leaves the same remainder as (xy)/7 would.
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Re: DS-Remainder of Quad Equation [#permalink] New post 02 Oct 2008, 21:42
Thats a new approach. I wasnt aware of it.
jatinrai wrote:
vr4indian wrote:
can anyone explain this?

IMO A
1) t divided by 7 leaves remainder 6.
t^2 divided by 7 leaves remainder same as 36/7 & i.e. 1
5t divided by 7 leaves remainder as 5*6/7 i.e. 2
6 divided by 7 leaves remainder as 6.
All 3 plus leaves remainder same as (1+2+6)/7 or 9/7 & i.e. 2

2) With this we cant calculate remainder for 5t. Hence insufficient.

Logic used: If X divided by A leaves remainder as x & Y divided by A leaves remainder as y, then X+Y when divided by A leaves the same remainder as (x+y)/7 would.
Also, XY when divided by A leaves the same remainder as (xy)/7 would.
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Re: DS-Remainder of Quad Equation [#permalink] New post 02 Oct 2008, 23:28
scthakur wrote:
From stmt1, t = 7n + 6 for n = 0,1,2,.....

Hence, t^2 + 5t + 6 = 49n^2 + 119n + 72
When you divide this expression by 7, the remainder will be 2. Hence, sufficient.

For stmt2, t^2 = 7n + 1 for n = 0,1,2,3....
Hence, t = sqrt(7n+1) and since t is an integer, hence, it can only be 1,5,9,....

For t = 1, the given expression will have remainder as 5.
For t = 5, the given expression will have remainder as 0.

Hence ,stmt2 is not sufficient.


It's a easiest way to solve 1, but not 2

In stmt2, t^2 + 7 divisible by 7

leaves 5t - 1

t = 2 => remainder 2
t = 3 => remainder 0
Re: DS-Remainder of Quad Equation   [#permalink] 02 Oct 2008, 23:28
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