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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
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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12 Jul 2009, 09:20
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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..
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.. 
12 Jul 2009, 16:36
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treemonkey
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.
