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# A Math Problem Soving questions

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Intern
Joined: 12 Jun 2009
Posts: 23
A Math Problem Soving questions [#permalink]

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12 Jul 2009, 02:34
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Hi,

Does anyone know why the correct answer to the below question is c -- 1.

If (a to the power of 2) - (b to the power of 2) = (c to the power of 2), which of the following CANNOT be a possible value of a + b + c

a. 4
b. 2
c. 1
d. 0
e. -2

Sorry I can't figure out how to type powers in this editor box.
This question has really puzzled me and any feedback is appreciated!!

Monkey
Intern
Joined: 11 Jul 2009
Posts: 7
Location: Delhi
Re: A Math Problem Soving questions [#permalink]

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12 Jul 2009, 08:20
Hi ,

You can always solve these problems using Hit and Trial.
You know a^2 = b^2 + c ^2
Now consider, a = 5, b = 4 , c =3
5^2 = 3 ^2 + 4 ^2 ( Pythagorus Theorm )
Now you can replace a as -5 b = 4 c =3 ( you get a + b + c = 2)
Replace a = 5 , b = -4 , c = -3 ( you get a + b + c = -2)
Replace a = 5 , b = -4 , c = 3 ( you get a + b + c = 4)

Also, a= 15 , b = -12 , c = -13 ( you get a + b + c = 0)

Although, the solution seems to be long.. but it takes hardly 1.5 min to solve..
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346
Re: A Math Problem Soving questions [#permalink]

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12 Jul 2009, 15:36
1
KUDOS
Expert's post
treemonkey wrote:
Hi,

Does anyone know why the correct answer to the below question is c -- 1.

If (a to the power of 2) - (b to the power of 2) = (c to the power of 2), which of the following CANNOT be a possible value of a + b + c

a. 4
b. 2
c. 1
d. 0
e. -2

They can all be values of a + b + c. For example, let a = 1/2, b = 0, and c = 1/2. Then a^2 - b^2 = c^2, and a + b + c = 1.

I'm guessing the question intends the numbers to be integers (if it doesn't say this, I'd be interested to know the source of the question). In that case, we could have:

* a and b are both odd, or both even, in which case c will be even, and a + b + c will be even;
* one of a or b is even, the other odd, in which case c will be odd, and a + b + c will be even.

So a + b + c cannot be odd if all three letters represent integers, and thus cannot equal 1.
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Intern
Joined: 12 Jun 2009
Posts: 23
Re: A Math Problem Soving questions [#permalink]

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14 Jul 2009, 01:36
Thank you so much guys. There's nothing like finally knowing the answer to a question that you've been working on for days!!!
Intern
Joined: 29 Jun 2008
Posts: 15
Re: A Math Problem Soving questions [#permalink]

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14 Jul 2009, 18:37
Hi All,

Nice question but i dont understand the last part of a+b+c as shown below:

Also, a= 15 , b = -12 , c = -13 ( you get a + b + c = 0)

15 - 12 - 13 is not zero rather it is -10; correct me if i am wrong.

ABAN
Intern
Joined: 12 Jun 2009
Posts: 23
Re: A Math Problem Soving questions [#permalink]

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15 Jul 2009, 03:12
Pareekpuneet,
No you are not wrong, I noticed that. I guess IanStewart's approach is better in this case. However it was good to do a revision on Pythagorus Theorm. I'm just grateful so many ppl are trying to help out..
Re: A Math Problem Soving questions   [#permalink] 15 Jul 2009, 03:12
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