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A Math Problem Soving questions

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New post 12 Jul 2009, 03:34
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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

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Re: A Math Problem Soving questions [#permalink]

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New post 12 Jul 2009, 09:20
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.. :lol: but it takes hardly 1.5 min to solve.. :lol:

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Re: A Math Problem Soving questions [#permalink]

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New post 12 Jul 2009, 16:36
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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.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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Re: A Math Problem Soving questions [#permalink]

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New post 14 Jul 2009, 02:36
Thank you so much guys. There's nothing like finally knowing the answer to a question that you've been working on for days!!!

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Re: A Math Problem Soving questions [#permalink]

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New post 14 Jul 2009, 19:37
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.

Thanks in advance.
ABAN

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Re: A Math Problem Soving questions [#permalink]

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New post 15 Jul 2009, 04:12
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..

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Re: A Math Problem Soving questions   [#permalink] 15 Jul 2009, 04:12
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