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I divided 4.18 by 5 by hand, like this:Division by Hand

As you see I removed the decimal point since I was dividing by a whole number, then I put the decimal point back and got 0.83 as result.

It's correct, the point is that often on this kind of division I get a remainder (in this case is 3), and I suppose this isn't right.

By the way, a calculator returns 0.836 for 4.18/5, which means I'm missing at least one decimal place.

My question is, how do I solve division with decimal points so that I get all the decimal places. Is it ok to have a remainder?

Thanks

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\begin{array}{cccccccccc} & & 0 & . & 8 & 3 & 6 \\ \\ 5 & ) & 4 & . & 1 & 8 & 0 \\ & & 0 \\ \\ & & 4 & & 1 \\ & & 4 & & 0 \\ \\ & & & & 1 & 8 \\ & & & & 1 & 5 \\ \\ & & & & & 3 & 0 \\ & & & & & 3 & 0 \\ \\ & & & & & & 0 \end{array}

The algorithm stops there because $0$ appeared as the remainder.

Where you have $4.18$ you can write $4.1800000000\ldots$ with as many $0$s as necessary.

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  • $\begingroup$ Hm... Interesting, now I can see it clearly. Let me ask just one more thing, when I remove the decimal, this means I'm multiplying the dividend (by 100 in this case). Wouldn't I have to immediately put a zero on the divisor? If so, where exactly? I mean, you already answered me with a good answer, but I wanna avoid mistakes, it's been a long time since the 5th grade now... $\endgroup$ – Ezequiel Barbosa May 19 '15 at 23:16
  • $\begingroup$ I'm not sure what you mean. You can say that $4.18\div 5=418\div500$. Is that what you meant? ${}\qquad{}$ $\endgroup$ – Michael Hardy May 20 '15 at 3:13
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By removing the decimal place, you are essentially computing $$\frac{418}{5},$$ which is exactly $100$ times your desired result.

In your computation, you obtain $$\frac{418}{5} = 83\ R3 = 83 + \frac{3}{5}.$$

This is equal to $83.6$; since this result is 100 times your desired result, divide by 100 to get

$$\frac{4.18}{5} = \frac{1}{100} \times \frac{418}{5} = \frac{1}{100}\times 83.6 = 0.836.$$

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The trick to these is to change the number so that it is easier to use, but has the same value. notice that $4.18 = 4.180=4.1800=\cdots$ so you can always add another $0$ onto the end of a decimal number and it will retain its value. In this case if you add a $0$ you will bring it down to the $3$ and need to find how many $5$'s go into $30$ and the answer is $6$. For a general problem, you will need to keep doing this until either you do not get a remainder, such as in this case, or you see a pattern in the decimals, in a problem such as $1÷3 = .333333333\ldots$ In the second case we can write the answer as simply $. \bar{3}$

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