Convert from base $10$ to base $5$ I am having a problem converting $727$(base $10$) to base $5$. What is the algorithm to do it?
I am getting the same number when doing so: $7\times 10^2 + 2\times10^1+7\times10^0 = 727$, nothing changes.
Help me figure it out!
 A: You need to count in terms of $5^0, 5^1, 5^2, 5^3$ and $5^4$, not in term of $100 ,10$ and $1$. 
Start by the highest power of $5$ smaller than your number, i.e. $5^4=625$ here, and check how much you can multiply it without exceding (this will be a number in $1,2,3,4$).
Take the leftover (base 10, 727-625=102) and repeat until you reach $0$. 
You should have
$$
\begin{align}
727_{10}&= 625+100+2\\
&= 1\times 5^4+4\times 5^2+2\times 5^0 \\
&= 10402_5.
\end{align}
$$
A: Trick: divide consecutively the quotients by $\;5\;$ and keep aside the residues :
$$\begin{align*}\frac{727}5=145+\frac25\longrightarrow& \color{red}2\\
\frac{145}5=29\longrightarrow&\color{red}0\\
\frac{29}5=5+\frac45\longrightarrow&\color{red}4\\
\frac55=1\longrightarrow&\color{red}0\\
\frac15=\frac15\longrightarrow&\color{red}1\end{align*}$$
Now put them in inverse order and voila!
$$727_{10}=10402_5$$
Can you see why it works?
A: You have to keep dividing by $5$ and write all the remainders.
So 
$$727 = 5\cdot 145 + 2$$
$$145 = 5\cdot29 + 0$$
$$29 = 5\cdot5 + 4$$
$$5 = 5\cdot1 + 0$$
$$1 = 5\cdot0 + 1$$
Write all the remainders in reverse: $10402$ 
And ta-da! There is your answer :-)
A: The trick is to realize that
\begin{align*}
727 &= 625 + 0*125 + 4*25 + 0*5 + 2 
\\ &= 5^4 + 0 *5^3 + 4*5^2 + 0*5^1 + 2*5^0.
\end{align*}
So the answer is $10402$.
A: You have to divide each number by 5 :
$$\begin{aligned}
727 &= 5 \cdot 145 &+ \color{red}{2}\\
145 &= 5 \cdot 29 &+ \color{green}{0}\\
29 &= 5 \cdot 5 &+ \color{blue}{4}\\
5 &= 5 \cdot 1 &+ \color{magenta}{0}\\
1 &= 5 \cdot 0 &+ \color{brown}{1}
\end{aligned}
$$
Hence the answer is $\color{brown}{1}\color{magenta}{0}\color{blue}{4}\color{green}{0}\color{red}{2}_5 = 1 \cdot 5^4 + 0 \cdot 5^3 + 4 \cdot 5^2 + 0 \cdot 5^1 + 2 \cdot 5^0 = 727$
