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How many digits are there when $7.30\times10^{28}$ is expressed in ordinary numeral? I thought there should be $30$ digits, but I'm wrong, why?

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  • $\begingroup$ Number of digits of a natural number $n$ is $\lfloor log n\rfloor + 1$ . $\endgroup$ May 11, 2016 at 11:52

4 Answers 4

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The problem with counting the number of digits in $7.30 \times 10^{28}$ is that $10^{28}$ is such a large number. So try some smaller numbers and see what happens.

$7.30\times10^{2} = 7.30 \times 100 = 730$ - 3 digits
$7.30\times10^{3} = 7.30 \times 1000 = 7300$ - 4 digits
$7.30\times10^{4} = 7.30 \times 10000 = 73000$ - 5 digits
$7.30\times10^{5} = 7.30 \times 100000 = 730000$ - 6 digits

This does not really prove that $7.30 \times 10^{28}$ has $29$ digits, but it is very compelling evidence.

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A way to understand the $\times10^{28}$ in $7.30 \times 10^{28}$ is, "Move all the digits of the number $28$ places to the left."

When you do this, $28$ digits that were to the right of the decimal point in $7.30$ (the $3$, the $0$, and several implicit trailing zeros) move to the left of the decimal point. But there was already one digit to the left of the decimal point (the $7$ in $7.30$), so now we have a total of $29$ digits to the left of the decimal point. (An "$n$-digit number" generally is a number with $n$ digits to the left of the decimal point, starting with a non-zero digit.)

If the question is how many significant digits are in $7.30 \times 10^{28}$, I would say there are three, because that's the number of digits that are written in $7.30$. The number $7.3000 \times 10^{28}$ has the same value but has five significant digits. The application for this is that we're measuring some actual quantity of something, and the more precise our measurement is, the more digits we can write while being sure that they are correct. (The way I was taught this is, once you reach a digit whose value is uncertain, you write that digit and then stop.)


I also remember being taught to "move the decimal point to the right" in order to multiply a number by a positive power of $10$. Perhaps it was taught this way because it's easy to erase (or cross out) the decimal point of a number and write a new decimal point between a different pair of digits. But as noted in the comments, "moving all the digits left" is a better way to describe this.

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  • $\begingroup$ I don't think moving the decimal point builds understanding of place value. I concede that it is done in practice but I think for an explanation it should be avoided. $\endgroup$
    – Karl
    May 11, 2016 at 18:18
  • $\begingroup$ @Karl Possibly one should say we shift all the digits of a number $k$ places to the left rather than shifting the decimal point $k$ places to the right. Digit-shifting is less "artificial" than moving a decimal point, it seems to serve very well in grade-school arithmetic, and it's also literally used in some electronic computations on binary numbers. $\endgroup$
    – David K
    May 11, 2016 at 19:24
  • $\begingroup$ I would wholeheartedly agree with moving the digits to the left and is how I normally teach things. I just wanted to mention it I hope you don't mind :) $\endgroup$
    – Karl
    May 11, 2016 at 19:28
  • $\begingroup$ @Karl I could rewrite the answer in terms of moving the digits. Maybe I should. $\endgroup$
    – David K
    May 11, 2016 at 19:31
  • $\begingroup$ Up-voted for the edits. (+1) $\endgroup$
    – Karl
    May 12, 2016 at 12:02
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$7.30*10^{28}$ when written in this form gives $73.0 * 10^{27}$

Then now we have two digits 7,3 and then 27 zeros after that which is indicated by $10^{27}$

=73000000000000000000000000000.0

The numbers after the decimal point doesn't count because it is zero. No matter how many zeros you add after the decimal point it is still going to remain the same number.

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You should think it through with a smaller example.

What does $2 \times 10^3$ mean?

Well, you are counting in thousands ($10^3$) and you have 2 of them.

But, because of the way that the decimal system works we say that we have:

no hundreds ($10^2$) and no tens ($10^1$) and no units ($10^0$)

So there are there are 3 place value columns between the tens and thousands (easily countable) and one more for the units (the sneaky $10^0$)

In your example it relates to $28+1=29$ if I have interpreted the question correctly.

Note I have assumed that you are not considering tenths or hundredths etc. Also the $0$ are place holders if that matters to you. There are only two non zero digits which I assumed you weren't asking about.

Hope this helps.

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