i am having actually a difficulty converting $727$(base $10$) to basic $5$. What is the algorithm to carry out it?

I am gaining the very same number once doing so: $7 imes 10^2 + 2 imes10^1+7 imes10^0 = 727$, nothing changes.

You are watching: How to convert base 10 to base 5

Help me figure it out!


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Trick: divide consecutively the quotients by $;5;$ and also keep aside the residues :

$$eginalign*frac7275=145+frac25longrightarrow& colorred2\frac1455=29longrightarrow&colorred0\frac295=5+frac45longrightarrow&colorred4\frac55=1longrightarrow&colorred0\frac15=frac15longrightarrow&colorred1endalign*$$

Now placed them in inverse order and voila!

$$727_10=10402_5$$

Can you see why that works?


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You need to count in terms of $5^0, 5^1, 5^2, 5^3$ and $5^4$, no in term of $100 ,10$ and $1$.

Start by the highest power of $5$ smaller sized than your number, i.e. $5^4=625$ here, and check how much you have the right to 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 with $0$.

You need to have$$eginalign727_10&= 625+100+2\&= 1 imes 5^4+4 imes 5^2+2 imes 5^0 \&= 10402_5.endalign$$


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You have to keep splitting by $5$ and also write every the remainders.

So

$$727 = 5cdot 145 + 2$$$$145 = 5cdot29 + 0$$$$29 = 5cdot5 + 4$$$$5 = 5cdot1 + 0$$$$1 = 5cdot0 + 1$$

Write all the remainders in reverse: $10402$

And ta-da! over there is her answer :-)


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The cheat is to establish thateginalign*727 &= 625 + 0*125 + 4*25 + 0*5 + 2 \ &= 5^4 + 0 *5^3 + 4*5^2 + 0*5^1 + 2*5^0.endalign*So the prize is $10402$.


You need to divide each number by 5 :

$$eginaligned727 &= 5 cdot 145 &+ colorred2\145 &= 5 cdot 29 &+ colorgreen0\29 &= 5 cdot 5 &+ colorblue4\5 &= 5 cdot 1 &+ colormagenta0\1 &= 5 cdot 0 &+ colorbrown1endaligned$$

Hence the answer is $colorbrown1colormagenta0colorblue4colorgreen0colorred2_5 = 1 cdot 5^4 + 0 cdot 5^3 + 4 cdot 5^2 + 0 cdot 5^1 + 2 cdot 5^0 = 727$


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