So I started with $gcd(10n-1,5n+6)$ States that $d|(10n-1,5n+6)$ Which also implies that $d|10n-1$ and $d|5n+6$. Now I dont know how to show gcd of equals to 1 or 13.
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$$\gcd(a,b) = \gcd(a,2b-a) = \gcd(a,13) \in {1,13} $$ since $13$ is a prime number. If $n\equiv 4\pmod{13}$, $13\mid(10n-1)$. – Jack D'Aurizio Oct 26 '16 at 06:34
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So, $d$ must divide $$2(5n+6)-(10n-1)$$
The basic idea is to eliminate $n$ from the relation to find a constant as dividend.
lab bhattacharjee
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@Kiterunner, So $d$ must divide $13$. What are the positive divisors of $$13?$$ – lab bhattacharjee Oct 26 '16 at 06:30
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Woh Nice one,,Can you tell me how does one figure out what you have to do..Not this simplification but from 10n-1,5n+6 how do you get What has t be done? – Arya Oct 26 '16 at 06:36
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@Kiterunner, Look at the last line of the answer. See also : http://math.stackexchange.com/questions/1852554/find-all-the-numbers-n-such-that-frac12n-610n-3-cant-be-reduced/1852765#1852765 – lab bhattacharjee Oct 26 '16 at 06:40
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Use that $(a,b)=(b, a-b).$
Then we have
$$ (10n-1,5n+6)=(5n+6,5n-7)=(5n-7,13)=13 \text{ or } =1. $$
Leox
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