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Given $a,b$ 2 numbers in $Z$ Where:

  1. $a$ is even number (ie there is $d$ in $Z$ S.T $a=2d$)
  2. $6a+11b=5$
  3. $a-b=5e$ for some $e$ in $Z$

My book claims: $gcd(2a+4b, 2a+8b)=20$

Why is that? How am I supposed to solve similar questions?

I know:

$a=2d$, $6a+11b=5$, $a-b=5e$

So:

$gcd(2a+4b, 2a+8b)=gcd(4(d+b),4(d+2a-10e)=4 gcd(d+b,d+2a-10e)$

I'm stuck here, where can I use the fact that $6a+11b=5$?

Bill Dubuque
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robert
  • 27

1 Answers1

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I think Bezout's identity[https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity] can be used to solve your problem.

The claim is that $gcd(b,d) = 5$ and the rest follows from the fact that $gcd(a,b) = gcd(b, a-b)$

From $6a+11b=5$ and $a-b=5e$ we see that 5 divides $d$ and $b$ and using Bezout's identity we say that $gcd(b,d) = 5$

Hope you will find this useful

Vasac
  • 73