Take any number. Drop the final two digits. Subtract from it nine times the number made by the two digits you dropped. The original number and the new number are either both divisible by 17 or both not divisible by 17.
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1can you illustrate by an example. – hamam_Abdallah Jun 19 '17 at 20:50
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2Hint: generalize to showing for any two integers $a, b$, $17 \mid 100a+b \Leftrightarrow 17 \mid a-9b$. Use arithmetic $\pmod{17}$. – Daniel Schepler Jun 19 '17 at 20:52
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Write your initial number $n$ as $n=100a+b$ where $b$ is the last two digits. You are then claiming that $100a+b \equiv 0 \pmod {17} \iff a-9b \equiv 0 \pmod {17}$ We have $100a+b \equiv -2a+b\pmod {17}$ Then if $-2a+b\equiv 0 \pmod {17}$ we can multiply it by $8$ and get $-16a+8b\equiv a-9b \equiv 0 \pmod {17}$ To go the other way we multiply $a-9b\equiv 0 \pmod {17}$ by $-2$ and retrace the steps.
Ross Millikan
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So your original number is $100a+b$. Then your new number is $a-9b$. Their difference is $99a+10b=6\times17a-3a-7b+17b\equiv-3a-7b\pmod{17}$. But, since $-3$ and $17$ are coprime,\begin{align*}a-9b\equiv0\pmod{17}&\Longleftrightarrow-3a+27b\equiv0\pmod{17}\\&\Longleftrightarrow-3a-7b\equiv0\pmod{17}\text,\end{align*}because $27\equiv-7\pmod{17}.$
José Carlos Santos
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