45 base 10 to base 5. I am getting problem problems arriving at an answer.

Question

I am getting problems with this equation

45 base 10 to base 5 I have tried this method 45÷5=9 r 0 9÷5=1 r 4 4÷5= 0 r 4 Ans =440. However when we do the reverse we are not getting 45. Someone please help

Answer 1

To convert you divide, keep track of the remainder and then divide the quotient and repeat until you get a quotient of $0$.

In you last step you confused a quotient of $1$ with the remainder of $4$.

So first step is $45 \div 5 = 9r 0$ so write down the $0$. You did that.

Then deal with the quotient. $9 \div 5 = 1r 4$ so write down the $4$ to get $40$. You did that.

Then deal with the quotient. The quotient is $1$. You did not deal with the quotient. You dealt with the remainder of $4$. That was wrong.

$1 \div 5 = 0 r 1$. So write down the $1$ to get $140$. And so $140_5 = 45_{10}$ and indeed $1*5^2 + 4*5 + 0*1 = 25 + 20 + 0 = 45$.

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To keep from getting confused it might help to add a "so that means..." after ever line:

$45\div 5 = 9 r0$:

so that means $45 = 9*5 + 0$.

And $9 \div 5 = 1 r 4$

so that means $9 = 1*5 + 4$

and that means $45 = 9*5 + 0 = (1*5 + 4)*5 + 0 = 1*5^2 + 4*5 + 0$

And $4 \div 5 = 0 r 4$

so that means $4 = 0*5 + 4$

and that means $45= 1*5^2 + 4*5 + 0$ and ... hey! That's a $1$, not a $4$!!!

So, ...oops...

And $1\div 5 = 0 r 1$

and that means $1 = 0*5 + 1$ so

$45 = (0*5 + 1)*5^2 + 4*5 + 0$ so

$45 = 0*5^3 + 1*5^2 + 4*5 + 0$ and we are done.

$45 = 1*5^2 + 4*5 + 0=140_5$