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Manager
Joined: 21 May 2008
Posts: 85

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28 Sep 2008, 00:49
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DS Question.
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DS-Remainder of Quad Eqn.doc [63.5 KiB]

Director
Joined: 23 Sep 2007
Posts: 783

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29 Sep 2008, 01:21
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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untitled.JPG [ 12.25 KiB | Viewed 881 times ]

Manager
Joined: 14 Jun 2007
Posts: 168
Location: Vienna, Austria

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30 Sep 2008, 02:34
SVP
Joined: 17 Jun 2008
Posts: 1547

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30 Sep 2008, 02:47
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.
Manager
Joined: 27 Aug 2008
Posts: 147

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02 Oct 2008, 10:10
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.
VP
Joined: 18 May 2008
Posts: 1260

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02 Oct 2008, 22:42
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.
Manager
Joined: 30 Sep 2008
Posts: 111

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03 Oct 2008, 00:28
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
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