## Switching Bases

For students of class 11-12 (age 16+)
sowmitra
Posts: 155
Joined: Tue Mar 20, 2012 12:55 am
Location: Mirpur, Dhaka, Bangladesh

### Switching Bases

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 $f_3(n)=8.n$?
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

kfoozminus
Posts: 33
Joined: Mon Nov 26, 2012 4:52 pm
Contact:

### Re: Switching Bases

$0$

Posts: 1016
Joined: Tue Nov 22, 2011 7:49 pm
Location: 127.0.0.1
Contact:

### Re: Switching Bases

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 $14+15=29$
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

kfoozminus
Posts: 33
Joined: Mon Nov 26, 2012 4:52 pm
Contact:

### Re: Switching Bases

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