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If x=6m^2+4n^2, and gcd
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06 Jul 2021, 07:53
06 Jul 2021, 07:53
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27% (01:06) correct
72% (01:05) wrong based on 11 sessions
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If \(x=6m^2+4n^2\), and \(gcd(m,n)=2\), what is the greatest even number that must be a factor of x ?
Show: :: OA
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Re: If x=6m^2+4n^2, and gcd
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07 Jul 2021, 02:39
07 Jul 2021, 02:39
2
Carcass wrote:
If \(x=6m^2+4n^2\), and \(gcd(m,n)=2\), what is the greatest even number that must be a factor of x ?
Show: :: OA
8
Let, \(m = 2x\) and \(n = 2y\) with \(GCD = 2\)
\(x = 2[3m^2 + 2n^2]\)
\(x = 2[3(2x)^2 + 2(2y)^2]\)
\(x = 2[12x + 8y]\)
\(x = 8[3x + 2y]\)
Hence, greatest even number that must be a factor of \(x\) is \(8\)
Re: If x=6m^2+4n^2, and gcd
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16 Aug 2021, 08:48
16 Aug 2021, 08:48
Why did we write it as 2x and 2y?
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