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For integer numbers $\overline{abcd}$ and $\overline{cdab}$ find great common divisor of $\overline{abcd}$ and $\overline{cdab}$

I tried $$\overline{abcd} = 100\overline{ab}+\overline{cd} \\ \overline{cdab} =100\overline{cd} +\overline{ab}\\ \Longrightarrow \overline{abcd}-\overline{cdab}=99(\overline{ab}-\overline{cd}).$$

I think $\gcd(abcd,cdab)=99$.

But then I find $\gcd(9680, 8096)=176$.

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  • $\begingroup$ I think $(ab - cd)$ should be $(\overline{ab} - \overline{cd})$. $\endgroup$
    – soupless
    May 9 at 15:25
  • $\begingroup$ $1300-13$ Is indeed divisible by $99.$ But neither $1300$ nor $13$ is divisible by $99.$ $\endgroup$ May 9 at 15:28
  • $\begingroup$ I suppose $\gcd(9999,9999)=9999$ is not allowed, but what is allowed? Can we have $a=c$ but $b≠d$? $\endgroup$
    – MJD
    May 9 at 15:31
  • $\begingroup$ The simple fact that $x-y=uv$ certainly doesn't imply that $u$ divides $x$ and $y$! Just think about that for a bit. $\endgroup$
    – TonyK
    May 9 at 15:33
  • $\begingroup$ The conclusion you should reach from $abcd -cdab = 99(ab-cd)$ is that $\gcd(abcd,cdab) = \gcd(abcd, [99(ab - cd)])= \gcd(abcd, 99)\gcd(abcd,ab-cd)$. It seems implied, but not stated, the $ab,cd$ are relatively prime so $\gcd(abcd,ab-cd)$ is implied to be $1$ (Although it need not be). But the $99$.... it is does not need to have any factors in common with $abcd$ or with $cdab$. $\endgroup$
    – fleablood
    May 9 at 15:56
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It is indeed true that the $abcd-cdab = 99(ab-cd)$.

ANd that does indeed mean that the $\gcd(abcd,cdab)$ does divide $99(ab-cd)$,

But that in no way means that $99$ divides or is or has any factors in common with $\gcd(abcd,cdab)$. (It doesn't mean it doesnt but it doesn't mean it does).

Indeed $9680 - 8096 = 99\cdot (96-80)$. And that means that all the factors $9680$ and $8096$ have in common are also factors of $99\cdot(96-80)$. But it doesn't mean that all of those factors are factors of $99$-- some of the factors might be factors of $96-80$. Nor does it mean that all the factors (or indeed any of the factors) of $99$ are common factors are $9680$ and $8096$.

What is does mean is that some distributed between the $99$ and the $96-80$ all the factors in common, and the $\gcd$ are to be found and indeed. $11|99$ and $16 = 96-80$ and the $\gcd(9680,8096) = 176 = 11\cdot 16$.

The conclusion you should reach is:

$\gcd(abcd,cdab) = \gcd(abcd, abcd-cdab)=$

$\gcd (abcd, 99(ab-cd))=$

$\gcd(99,abcd)\cdot \gcd(abcd, ab-cd)=$

etc.

$\gcd(9680-8096) = \gcd(9680,9680-8096)=$

$\gcd(9680, 99\cdot 16) = $

$\gcd(9680, 99) \gcd(9680, 16)=$

$\gcd(9680- 96\cdot 99, 99)\gcd(9680-1600, 16)=$

$\gcd(176,99)\gcd(8080,16)=$

$\gcd(176-99, 99) \gcd(80\times 101, 16) = $

$\gcd(77,99)\gcd(80,16)\gcd(101,16)=$

$\gcd(77, 99-77)\gcd(5\times 16, 16)\gcd(101-6*16,16)=$

$\gcd(77,22)\cdot 16\cdot(5,16) =$

$\gcd(77-3\cdot 22, 22)\cdot 16\cdot \gcd(16-3\cdot 5, 5)=$

$\gcd(11,22)\cdot 16\cdot \gcd(1,5) =$

$\gcd(11, 2*11)\cdot 16 \cdot 1=$

$11 \cdot 16 \cdot 1 = 176$.

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We'll assume $\overline{ab} \ne \overline{cd}$, since the solution is obvious in that case.

Let $n = \overline{ab}/GCD(\overline{ab}, \overline{cd})$, $m = \overline{cd}/GCD(\overline{ab}, \overline{cd})$, and $g = GCD(100m + n,100n +m)$. Clearly $m$ and $n$ are relatively prime and $GCD(\overline{abcd}, \overline{cdab}) = GCD(\overline{ab}, \overline{cd})g$. We also have that $g = GCD(100m + n,99(n-m)) = GCD(100m+n, 101(m+n))$, so $g | 99(n-m)$ and $g | 101(n+m)$. We can't possibly have $g | 101$, and $101$ is prime, so $g | m + n$. That means that $g$ and $n - m$ are relatively prime, since $m$ and $n$ are relatively prime, and so $g | 99$. Further, any number that divides both $99$ and $m+n$ also divides $g$, since it would divide both $99(n-m)$ and $99m + m + n$. So $g = GCD(m + n, 99)$.

Thus, we have $$ GCD(\overline{abcd}, \overline{cdab}) = GCD(\overline{ab}, \overline{cd})GCD\left(\frac{\overline{ab} + \overline{cd}}{GCD(\overline{ab}, \overline{cd})}, 99\right), $$ or equivalently, $$ GCD(\overline{abcd}, \overline{cdab}) = GCD\left(\overline{ab} + \overline{cd}, 99\cdot GCD(\overline{ab}, \overline{cd})\right) $$

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