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If $5x =7\pmod{13}$ and $3y =10\pmod{13}$. What is the value of $k$ in $xy =k\pmod{13}$

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Hint $\rm\,\ mod\ 13\!:\,\ \color{#C00}7\equiv \color{#0A0}{-6}\ \Rightarrow\ x\cdot y\, \equiv\, \dfrac{\color{#C00}7}{5}\cdot\dfrac{10}3\, \equiv\, \dfrac{\color{#0A0}{-6}}5\cdot \dfrac{10}3\,\equiv\, -2\cdot 2$

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We have that $(5x)(3y)=5 \pmod{13}$. As well $5\cdot 3=2\pmod{13}$. So we just need to find the inverse of $2$ under mod 13. This is easily seen to be 7. Hence,

$xy=5\cdot 7\pmod{13}=9\pmod{13}$

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this is not clear to me, why not divide $(5x)(3y)=5\pmod{13}$ by 5 then look for the inverse of 3 instead. –  Vaolter Oct 25 '12 at 16:04
    
Sure that works too. I was trying to show it more generally. –  Alex R. Oct 25 '12 at 16:18
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