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If a, b and c are even integers, which of the following coul

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rk21857
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If a, b and c are even integers, which of the following coul [#permalink] New post   08 Mar 2011, 12:03
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If a, b and c are even integers, which of the following could be the value of a^2 + b^2 + c^2,

a) 140
b) 246
c) 638
d) 862
e) 118
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A
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Bunuel
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Re: Sum of Square of Even intergers [#permalink] New post   08 Mar 2011, 12:20
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rk21857 wrote:
If a, b and c are even integers, which of the following could be the value of a^2 + b^2 + c^2,

a) 140
b) 246
c) 638
d) 862
e) 118

I am not sure about the answer, I just picked 246.


As a, b and c are even integers then a^2+b^2+c^2=(2x)^2+(2y)^2+(2z)^2=4(x^2+y^2+z^2), so the sum must be a multiple of 4 (note that a number is divisible by 4 if the last two digits form a number divisible by 4), the only choice which is a multiple of 4 is A: 140=10^2+6^2+2^2.

Answer: A.
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geisends
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Re: Sum of Square of Even intergers [#permalink] New post   08 Mar 2011, 13:48
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Can you explain the part where you take the 4 out of the equation in layman's terms. please
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Re: Sum of Square of Even intergers [#permalink] New post   08 Mar 2011, 13:53
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geisends wrote:
Can you explain the part where you take the 4 out of the equation in layman's terms. please


Just factor out 4 out of \(4x^2+4y^2+4z^2\): \((2x)^2+(2y)^2+(2z)^2=4x^2+4y^2+4z^2=4*(x^2+y^2+z^2)\).
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Re: Sum of Square of Even intergers [#permalink] New post   08 Mar 2011, 14:19
4x^2+4y^2+4z^2 : How do you know which number to manipulate it to. How do you know not to put a 6 or 8 as the coefficient instead of 4, or whether to leave it as 2 or change it to something else.
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Re: Sum of Square of Even intergers [#permalink] New post   08 Mar 2011, 14:25
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geisends wrote:
4x^2+4y^2+4z^2 : How do you know which number to manipulate it to. How do you know not to put a 6 or 8 as the coefficient instead of 4, or whether to leave it as 2 or change it to something else.


Please read the stem: "If a, b and c are even integers..." So a, b and c can be expressed as a=2x, b=2y, and c=2z for some integers x, y, and z --> a^2+b^2+c^2=(2x)^2+(2y)^2+(2z)^2=4(x^2+y^2+z^2).
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Re: Sum of Square of Even intergers [#permalink] New post   08 Mar 2011, 14:29
why does 2x)^2+(2y)^2+(2z)^2=4(x^2+y^2+z^2)? If you multiply 4 by each variable wouldnt that make 4x^2+4y^2+4z^2. I read the question stem Im just not getting it....
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Re: Sum of Square of Even intergers [#permalink] New post   08 Mar 2011, 14:48
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geisends wrote:
why does 2x)^2+(2y)^2+(2z)^2=4(x^2+y^2+z^2)? If you multiply 4 by each variable wouldnt that make 4x^2+4y^2+4z^2. I read the question stem Im just not getting it....


I don't quite understand your question but anyway, step by step:

\(a^2+b^2+c^2=(2x)^2+(2y)^2+(2z)^2=4x^2+4y^2+4z^2=4*(x^2+y^2+z^2)\).
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Re: Sum of Square of Even intergers [#permalink] New post   08 Mar 2011, 17:47
My answer is A.

there is atleast one two in every even integer,in other words even integer can be expressed as 2 times something

and squares will have atleast one 4 in them.

so answer should be a multiple of 4 , out of the given answers only A is multiple of 4.
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If a, b and c are even integers, which of the following coul [#permalink] New post   07 Nov 2019, 04:07
Although above solutions are more elegant than mine, you can also simply list some even perfect squares

\(4,16,36,64,100,144,256,324,400\)

then, through trial and error, \(4+36+100\) produces a match
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Re: If a, b and c are even integers, which of the following coul [#permalink] New post   02 Dec 2023, 04:59
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Re: If a, b and c are even integers, which of the following coul [#permalink]
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