Switching Bases

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sowmitra
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Switching Bases

Unread post by sowmitra » Wed Dec 19, 2012 9:42 pm

For all positive integers $n$, let $f_3(n)$ be the representation of $n$ in $Base-3$, considered as a number in $Base-10$. For example, since $5$ in $Base-3$ is $12_3$, so $f_3(5)=12$. What is the sum of all positive integers $n$ such that $f_3(n)=8.n$?
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kfoozminus
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Re: Switching Bases

Unread post by kfoozminus » Thu Dec 20, 2012 10:57 am

$0$ :?
$f_3(n)-n=7 \cdot \lfloor \frac{n}{3} \rfloor $

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Phlembac Adib Hasan
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Re: Switching Bases

Unread post by Phlembac Adib Hasan » Thu Dec 20, 2012 11:51 am

kfoozminus wrote:$0$ :?
$f_3(n)-n=7 \cdot \lfloor \frac{n}{3} \rfloor $
Check your thinking and explain to us.
There are such numbers and their sum is $29$.
Firstly notice that $f_3(n)$ is divisible by $8$. By checking the cases we see the last three digits of $n_3$ must be of the form $112_3$ or $200_3$. So the equation can be re-written as \[\displaystyle f_3\left (\frac {n-14}{27} \right )\times 125+14=n\]\[\boxed {OR}\]\[\displaystyle f_3\left (\frac {n-15}{27} \right )\times 125+15=n\]
Simple logic (actually the massive size of trinary numbers) shows that $\displaystyle f_3\left (\frac {n-14}{27} \right )\times 125>n$ and $\displaystyle f_3\left (\frac {n-15}{27} \right )\times 125>n$ for $n>14$ and $n>15$, respectively. Also the equations (above) are not valid for $n<14$. Now by checking we get $n=14,15$. So their sum is $14+15=29$
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kfoozminus
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Re: Switching Bases

Unread post by kfoozminus » Thu Dec 20, 2012 12:05 pm

sorry, my previous post actually doesn't make any sense. it was a mistake. :oops:

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