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# why this is wrong?

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29 Aug 2008, 09:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

where did i go wrong???
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Re: why this is wrong? [#permalink]

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29 Aug 2008, 09:51
arjtryarjtry wrote:
where did i go wrong???

_________________

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Current Student
Joined: 11 May 2008
Posts: 556
Followers: 8

Kudos [?]: 174 [0], given: 0

Re: why this is wrong? [#permalink]

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29 Aug 2008, 10:02
pls guide me how to do so ,
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Re: why this is wrong? [#permalink]

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29 Aug 2008, 10:03
arjtryarjtry wrote:
where did i go wrong???

A. Tough one.

(1) Sufficient
t = 7k+6, where k is an integer

(7k+6)^2 + 5(7k+6) + 6
= 49k^2 +84k + 36 + 35k + 30 + 6
= 49k^2 + 119k + 72

49k^2 is divisible by 7.
119k is divisible by 7.
Thus, remainder is always remainder of 72/7, or 2.

(2) Insufficient

t^2 = 7k+1
t = +/-sqrt(7k+1) = an integer
t could be 6, 9, 15 etc...

If t = 6
t^2 + 5t + 6 = 36 + 30 + 6 = 72
72/7 = 10 R 2

If t = 15
t^2 + 5t + 6 = 225 + 75 + 6 = 306
306/9 = 34

Last edited by zonk on 29 Aug 2008, 10:41, edited 3 times in total.
SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 34

Kudos [?]: 865 [0], given: 5

Re: why this is wrong? [#permalink]

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29 Aug 2008, 10:04
arjtryarjtry wrote:
where did i go wrong???

1)
t= 7k+6

t^2+5t+6= 49k^2+2*7k*6+36+6
=49k^2+2*7k*6+72
divide this by 7 remainder always 2
suffcient
2)
t^2=7m+1

t^2+5t+6= 7m+1 +5t+6= 7m+7+ 5*t

m=9 t=8 5t = 40 REMAINDER 5
m=32 t=15 5t = 120 REMAINDER 1

not suffcient

A
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Re: why this is wrong?   [#permalink] 29 Aug 2008, 10:04
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# why this is wrong?

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