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# x+1)(y-1) = 16, x > 0, y > 1 then 1) x + y >=

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CEO
Joined: 15 Aug 2003
Posts: 3452

Kudos [?]: 924 [0], given: 781

x+1)(y-1) = 16, x > 0, y > 1 then 1) x + y >= [#permalink]

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08 Sep 2003, 16:28
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(x+1)(y-1) = 16, x > 0, y > 1 then

1) x + y >= 8
2) x + y >= 4
3) x + y <= 8
4) x + y <= 4

my solution in yellow..and i can handle this in ok time...but as always..faster the better

For a good approximation...lets consider only integers

x ... y .....(x+1) (y-1).... Sum( x+y)
1 ... 9 .... 16 . ............ 10
3 ... 5 .... 16 ............ 8
7 ... 3 ... 16 ................ 10
15... 2 .... 16 ................ 17

These are all the possible values which will give us 16...
and the sum is alteast 8....

the answer is A...

thanks
praetorian

Kudos [?]: 924 [0], given: 781

CEO
Joined: 15 Aug 2003
Posts: 3452

Kudos [?]: 924 [0], given: 781

Re: PS [#permalink]

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13 Sep 2003, 05:53
praetorian123 wrote:
(x+1)(y-1) = 16, x > 0, y > 1 then

1) x + y >= 8
2) x + y >= 4
3) x + y <= 8
4) x + y <= 4

thanks
praetorian

stolyar

would you mind helping me out with a faster way?

thanks

Kudos [?]: 924 [0], given: 781

Intern
Joined: 06 Sep 2003
Posts: 19

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

Location: island

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13 Sep 2003, 09:01
(x+1)(y-1) = 16, x > 0, y > 1 then

1) x + y >= 8
2) x + y >= 4
3) x + y <= 8
4) x + y <= 4

(x+1)(y-1)=16

we can let (x+1)=4, (y-1)=4, got x=3 and y=5,that is the smallest answer for x and y, , so x+y>=8

thanks!!

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

13 Sep 2003, 09:01
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# x+1)(y-1) = 16, x > 0, y > 1 then 1) x + y >=

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