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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: 3470
Followers: 61

Kudos [?]: 697 [0], given: 781

x+1)(y-1) = 16, x > 0, y > 1 then 1) x + y >= [#permalink]  08 Sep 2003, 15:28
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 3 sessions
(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....

thanks
praetorian
CEO
Joined: 15 Aug 2003
Posts: 3470
Followers: 61

Kudos [?]: 697 [0], given: 781

Re: PS [#permalink]  13 Sep 2003, 04:53
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
Intern
Joined: 06 Sep 2003
Posts: 19
Location: island
Followers: 0

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

(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!!
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