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# Well, this one seems to be straightforward, but there are

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Director
Joined: 19 Mar 2007
Posts: 522

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Well, this one seems to be straightforward, but there are [#permalink]

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12 May 2007, 13:17
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Well, this one seems to be straightforward, but there are some doubts....
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Senior Manager
Joined: 03 May 2007
Posts: 270

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12 May 2007, 13:26
you need both statements to solve this one.
2 equations 2 unknowns

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Director
Joined: 19 Mar 2007
Posts: 522

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12 May 2007, 13:32
Sergey_is_cool wrote:
you need both statements to solve this one.
2 equations 2 unknowns

But why cannot we guess the values from st1? We've got two sqrds, the difference between which is 16, the first sqrd is greater than the second one: 5^2 - 3^2 = 16. Are there any other sqrds the difference between which can give us 16?

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VP
Joined: 08 Jun 2005
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12 May 2007, 13:42
I would say (C)

a^4-b^4 = (a^2-b^2)(a^2+b^2)

Statement 1

(a^2-b^2)=16

we can plug in but we are still missing (a^2+b^2)

insufficient

Statement 2

(a+b)=8

we are still missing (a^2+b^2)

insufficient

Statement 1&2

Knowing that (a^2-b^2)=(a-b)(a+b)=16

and (a+b)=8

we can plug in and find that (a-b)*8=16

a-b=2 & a+b=8

2b=6

b=3

a=5

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Senior Manager
Joined: 03 May 2007
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12 May 2007, 13:43
yeah but you can't just guess numbers you have to have the way to prove that the the equation has only one solution.(It's my guess) Do you have the answer?

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VP
Joined: 08 Jun 2005
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12 May 2007, 13:48
nick_sun wrote:
Sergey_is_cool wrote:
you need both statements to solve this one.
2 equations 2 unknowns

But why cannot we guess the values from st1? We've got two sqrds, the difference between which is 16, the first sqrd is greater than the second one: 5^2 - 3^2 = 16. Are there any other sqrds the difference between which can give us 16?

please note that a & b don't have to be integers !!!!!

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12 May 2007, 13:48
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