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In mathematics, as far as I know, you can't divide 100% by 3 without having 0,1...% left.

Imagine an apple which was cloned two times, so the other 2 are completely equal in 'quality'. The totality of the 3 apples is 100%. Now, you can divide those 3 apples for 3 persons and you will get 100% divided by 3 and none left.

Is this because 1: mathematics is not real 2: there is no 1 or 2, and it's in fact just an invention for measurements? So, in the division of 100% by 3 WITHOUT any left, is NOT accurate?

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    $\begingroup$ Why the downvotes? IMO, the question is well phrased: it starts with OP's own idea ("as far as I know ...") and tells an observation that seems to be in contradicition with that idea, and even gives some guesses what the reason might be. The question does not seem to be Google-trivial - searches like "dividing by three" or "is it possible to divide by three" seem to return completely unrelated results. If it is a duplicate, I think people should give a link to the older question instead of just downvoting. $\endgroup$
    – JiK
    Mar 30 '14 at 16:29
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    $\begingroup$ I am reminded of a similar puzzle. Is it possible to divide 20 sugar cubes amongst three cups of tea such that each cup has an odd number of sugar cubes, no cubes are divided, and none are left over? It seems impossible, but what you do is one sugar cube in the first cup - that's an odd number. One sugar cube in the second cup - that's an odd number. And 18 sugar cubes in the last cup, because 18 is surely an odd number of sugars to put in your tea. $\endgroup$ Mar 31 '14 at 15:26
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    $\begingroup$ More seriously: you seem to be focussed on "100%" as being special, but "100" is only special because of the accident of history that we have 10 fingers and so like to divide things into 10 or 10x10 parts. If we had 12 fingers and divided things into 12 or 144 parts then you'd have no problem thinking of a third as being a natural division. (And incidentally this is why we have 12 hours in a day, 60 minutes in an hour and 60 seconds in a minute: so that things divide more evenly more easily.) $\endgroup$ Mar 31 '14 at 15:30
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    $\begingroup$ @Anupam: You might not be a native English speaker and therefore not get the joke. "Odd" in English means both "a number which gives a remainder of one when divided by two" and "something unusual". 18 cubes is an unusual quantity of sugar to put in a cup of tea, I hope you agree. $\endgroup$ Mar 31 '14 at 15:31
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    $\begingroup$ @Anupam: It's a joke. The joke is that I use "odd" to mean "a non-even number" in the first two cups and "an unusual number" in the last one. That's why it's funny. Note that the function of my sense of humour is to amuse me; apparently I did not amuse you! $\endgroup$ Mar 31 '14 at 15:41

14 Answers 14

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So in mathematics, as far as I know, you can't divide 100% by 3, without having that 0,1..% left....

No! we can in Math and also in real life. This is similar to ask can we divide $1$ into 3 parts? And the answer is again yes. $$1÷3=\frac{1}{3}$$ because adding $\frac{1}{3}$ three times give $1$.

Consider 3 sticks of same length. Align these three sticks and call the total length as 1 unit. Now the length of any of the individual stick is exactly $\frac{1}{3}$ unit.

Moreover you can use the number system having base 3 to remove the (apparent) incompleteness of the base-ten expression $1÷3=0.333333333...$ In the number system having base 3, the number '3' itself will be written as $10$ and the number $1$ as it is.

The division $1÷3$ is now $1÷10$ which is equal to $0.1$. so you see writing (in base ten) $100÷3=33.333333$ does not mean that we cannot divide $100$ into three equal parts. What it means is that we are using a number system having $10$ base so we cannot write $\frac{100}{3}$ in decimals.

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    $\begingroup$ With regard to the Question, this can be applied to say one-third of $100$% is $33 \frac13$%. $\endgroup$
    – hardmath
    Mar 30 '14 at 11:47
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    $\begingroup$ In base $9$, $100_{10}=121_9$ so that $$\frac{100_{10}\%}{3}=\frac{121_9 \%}{3}=36.3_9 \%$$ You could also use base $3$. $\endgroup$
    – user5402
    Mar 30 '14 at 12:53
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    $\begingroup$ yeh that was my perspective, "-- 3 sticks of same length. Align these three sticks and call the total length as 1 unit. Now the length of any of the individual stick is exactly 1/3 unit." I thought that in mathematics there would always be 0.1% left. Guess not, thanks! $\endgroup$
    – RAO
    Mar 30 '14 at 13:33
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    $\begingroup$ @RAO: Another way of viewing this is that only a small bit of maths is about how you show a resulting number. The division operation, when and how to use it, and how it relates to other operations on numbers, is more fundamental than how you display those numbers. $\endgroup$ Mar 30 '14 at 15:18
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    $\begingroup$ @metacompactness - I really like your solution: The whole problem is simply with our decimal representation, not with the quantity itself. Change your radix to 3 and the problem disappears. $\endgroup$
    – Vector
    Mar 31 '14 at 6:07
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Meanwhile in ancient Greece...

For quite a long time, greek (and not only) mathematicians described numbers as lengths of certain line segments. So, when asked "What is $\sqrt{2}$ equal to?" they'd draw a $1\times1$ square (nevermind the unit), draw it's diagonal and say "There it is! This diagonal's length equals exactly $\sqrt{2}$!". So to answer your question: draw yourself a line, pick up a calliper, and divide this line 3 times. Like so:

dividing a line segment to 3 equal parts

And there you have it: 100% of a line segment divided into 3 equal parts. And if you ask "Yes, but what is this $\frac{1}{3}$ really equal to?" ancient philosopher would show you one of the parts and say "There it is! This segment's length equals exactly $\frac{1}{3}$!"

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    $\begingroup$ This is a great answer: The ancients used geometry. $\endgroup$
    – Vector
    Mar 31 '14 at 1:20
  • $\begingroup$ I don't really get it. How do you divide the line with the calliper? What's the use of 3 parallel red lines? $\endgroup$
    – hans-t
    Mar 31 '14 at 10:19
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    $\begingroup$ @user74158 The thing is, using this method, you can draw any line through the "end" (e.g. right-hand side) of the line segment L, set your calliper to any distance and draw n points of equal distance on that line. After connecting the nth point to the start of L, drawing n-1 lines parallel to the outer line in a way so they intersect the other points, you will get n equal segments on your line segment L by the laws of geometry. This solution allows you to draw a line at any angle (though somewhere around 90 will be easiest) and have any (equal) distance between your points $\endgroup$
    – Tim Meyer
    Mar 31 '14 at 10:55
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    $\begingroup$ Now, dividing an angle into three equal parts is a whole different ball game! $\endgroup$
    – user1729
    Mar 31 '14 at 21:36
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    $\begingroup$ @DigitalTrauma you can always use "Eureka!" if you get bored! Btw, I've never expected to get so many upvotes. A bit more and I'll exceed my reputation on Stackoverflow, and I posted 37 more answers there than here... $\endgroup$
    – Dunno
    Apr 2 '14 at 18:12
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One way to see this is to distinguish between definition and representation. I think you have began that.

  1. Definition: 1/3 is by definition exactly one third of 1.
  2. Representaion: "1/3" is a representation that is useful here. However there is no representaion of 1/3 using something like "0.3333333...". Unless you have an infinite paper or screen, of course. ;-)
  3. Another question, which you also asked, is mapping of the definitions to the real world. A very difficult question. However you could look at this from the aspect of usefulness. We measure something. That gives us a number. What can it be used for? How accurate is it? Will we get the same number the next time? Again a very difficult question - at least if we are outside the more obvious cases. This is today a big obstacle in science. (Read John P Ioannidis research to get touch of this.)

Good luck with your thoughts and investigations! :-)

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    $\begingroup$ +1 : to distinguish between definition and representation $\endgroup$
    – Vector
    Mar 31 '14 at 4:23
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    $\begingroup$ +1--but I would quibble with 0.3333333... not being an exact representation. It is exact, you just need to understand what it means. I don't need to write an infinite number of zeros at the end of 1.5 when I am representing 3/2 as a decimal. We just understand that I mean "no other negative powers of ten have coefficients here, ever, out to infinity". When you write .333... I assume you mean 3's repeating forever (better with an overbar, but still, it's clear what you mean, and it's mathematically exact). $\endgroup$
    – msouth
    Mar 31 '14 at 13:03
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    $\begingroup$ Yes, you are right, @msouth. "0.333..." could be seen as an exact definition. I was not careful enough with the definition. ;-) $\endgroup$
    – Leo
    Mar 31 '14 at 14:19
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    $\begingroup$ @Blackhawk Ah, but 0.999… is exactly 1. $\endgroup$
    – ghoppe
    Mar 31 '14 at 20:27
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    $\begingroup$ @Leo, your answer would be much better if you stopped trying to accommodate the possibility of "other definitions " and instead corrected your answer; as it stands, the statement "there is no representation of 1/3 using something like '0.3333333...'" is not just incorrect but self-contradictory, since 0.3333333... is just such a representation. $\endgroup$ Apr 3 '14 at 1:32
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The fundamental fallacy in your reasoning is that "natural == real". Just because a number never ends doesn't mean it isn't a "real" number with real application.

You have three apples. Three is a "natural number"; it is positive and whole, used in counting and other "everyday math".

Now, you are converting these three apples to "100%". "Percent" is from Latin per centum, literally meaning "per hundred", and so it defines a ratio with a common base; one hundred "per hundred" is everything; the whole. That's why the percent sign % looks the way it does; it's a symbol representing a fraction, that fraction here being the percentage quantity divided by 100.

Now, you have to ask (and the question doesn't), what is "the whole"? In this case, it's the three apples. You can't speak in percentages without speaking in percentage of the whole, requiring you to define the whole. 100% of three apples is three apples, and that's all there is to it.

Now, one-third of 100 is 33.3333... It is not a "whole", "natural" number, because 100 cannot be divided "evenly" by 3. This is utilized in the problem to offend our sensibilities, because one apple is a whole thing, and that's one-third of 100% of the apples. However, 100(%) means three apples; it's one hundred out of (divided by) one hundred of (times) three apples, divided by 3. The math, therefore, holds:

$$\dfrac{(\dfrac{100}{100} * 3)}{3} = 1$$

The question essentially intends to trick us by basically stating 100=3 and hoping we won't notice the switcheroo.

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    $\begingroup$ This is the most direct answer to the original question, making it the best in my opinion. $\endgroup$
    – Jace
    Mar 31 '14 at 21:43
  • $\begingroup$ '+1'. An alternative but similar explanation of this answer is: Let the three apples weigh 100 schrams(schram is a unit we made like gram). Now 100 schrams means three apples so one apple is 33.333...schrams but still one apple is a whole in itself. 100 schrams = 3 apples. $\endgroup$
    – user103816
    Apr 1 '14 at 5:55
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It's been answered above, but I thought I should try for a consise answer.

The fact that 100, in base 10, divided by 3 lacks a finite representation does not mean that it's impossible to divide 100 by 3.

In fact most real numbers can not be represented by a finite decimal number.

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  • $\begingroup$ In fact there's infinitely more of those, as in $\mathfrak{c}$ instead of $\aleph_0$ $\endgroup$
    – Dunno
    Mar 31 '14 at 13:26
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There are a number of conceptual errors in your question.

In mathematics, as far as I know, you can't divide 100% by 3 without having 0,1...% left.

In mathematics, "100%" means nothing more or less than "100 per 100", namely "$100 / 100 = 1$. So in mathematics you can divide $100\%$ by $3$ without having $0.1\%$ left. $100\% / 3 = 1 / 3 = \frac13$.

Imagine an apple which was cloned two times, so the other 2 are completely equal in 'quality'. The totality of the 3 apples is 100%. Now, you can divide those 3 apples for 3 persons and you will get 100% divided by 3 and none left.

Here you correctly observe that if you consider a group of 3 identical apples, and count those apples as 100%, then you can divide that group up into 3 (equal) parts and get none left. As explained above, this is not contradictory to "dividing 100% by 3".

Is this because 1: mathematics is not real 2: there is no 1 or 2, and it's in fact just an invention for measurements? So, in the division of 100% by 3 WITHOUT any left, is NOT accurate?

No.

The mathematics concerning integers and rationals and real numbers is very real, in the sense that using purely logical reasoning we can from mathematical axioms prove rigorously and unambiguously many statements about integers and rationals and reals that can be interpreted as statements about real-world phenomena, and in such a way that these statements can be empirically verified. More specifically, this part of mathematics has both explanatory power and predictive power about the real world, meaning that it provides succinct explanation for phenomena that we have observed and that it allows us to predict future phenomena that we have not yet observed.

For instance, right now the webpages you are reading from https://math.stackexchange.com have been encrypted using RSA encryption, and the decryption process relies on Fermat's little theorem to be true when interpreted as a statement about large positive integers encoded in your computer in binary format (with the addition and multiplication operations defined according to how your computer does them)!

In other words: If Fermat's little theorem (interpreted as a predictive statement about your computer) fails to be true, then you simply would be unable to read these webpages!

It turns out that Fermat's little theorem can be proven in a suitable form in a mathematical theory called PA (First-Order Peano Arithmetic). And so far, every single known theorem of PA has not been found false under the standard interpretation as a real-world statement. So we definitely have good reason to believe that "mathematics generated by PA is real".

As for $1$ and $2$ being inventions, the abstract concept of the natural numbers as a model of PA (i.e. they satisfy all the axioms of PA) is indeed a human invention, but that does not necessarily imply that this abstract concept has no real-world meaning, as explained above.

To make it clear, 100% can be divided exactly by 3, both in mathematics and in a suitable interpretation of the concepts "100%" and "3" in the real-world.

~ ~ ~

Coming back to your mention of "$0.1\%$", it is actually due to a conceptual mistake. It is a (mathematical) fact that $100$ is not an integer multiple of $3$. That is, there is no integer $k$ such that $100 = 3 × k$. However, there is a rational $r$ such that $100 = 3 × r$. When we write "$100\% / 3$" in mathematics, we mean exact division, which in this case yields the rational number $\frac13$ as the answer.

If you punch a basic calculator (which is designed to show you only some numbers in base ten) and ask for 100% / 3, then you may get something like 0.333333333. Why? It is not because the answer is $0.333333333$. Rather, the calculator is unable to show the answer in base ten. Furthermore, if you set your calculator to round the answer to 3 decimal places, then you would instead get 0.333, only because you asked for the answer rounded to 3 decimal places! So the (mathematical) fact that $100\% = 0.333 × 3 + 0.1\%$ does not actually have anything to do with the division of $1$ by $3$.

By the way, this phenomenon involving calculators is an artifact of our choice of base system for general daily life, which influences the design of such calculators. This has nothing to do with the mathematical concept of division. If humans had only 3 fingers on each hand, we might have ended up using base six for daily life.

We can in fact show precisely what would have happened if we used base $6$. We (using base ten) write $21_6$ to mean the number represented by the string of symbols "21" in base $6$, which means $2 × 6 + 1 = 13$. Likewise, in base $6$ we would have $1 / 3 = 0.2_6$ because $\frac13 = \frac26$. Note that we also have $\frac13 = 0.333\overline{3}$, where the overlined "$3$" means that "$3$" is repeated forever. There is no contradiction here. Whenever we do not write the subscript on a number, it is by convention understood as a base ten number. To make clear, $\frac13 = 0.333\overline{3}_{10} = 0.2_6$.

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In the real numbers, you can divide any quantity by 3 and get 3 equal parts. This is because of how we've defined the real numbers.

You pose a very valid question, though, as real numbers aren't the only way to model quantities. For instance, we could also use the natural numbers (the ones we use for counting things discretely.) For instance, if you were to consider all of the atoms present in an apple, the apple would only be divisible into 3 parts with the same number of atoms if the total count were a multiple of 3. With that choice of representation, it's true that it may not be possible to divide an apple into three equal parts.


At the risk of creating confusion (spurring curiosity?), here is another interesting conundrum. If you're sufficiently creative in how you cut up the apple, you can actually create two apples of identical volume from the single apple. This follows from certain assumptions we'd really like to be true, but it's certainly a counter-intuitive result. This begs us to question what we're taking for granted in our intuitive understanding and, indeed, whether our intuition is actually self-consistent.

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The Apple LOOKS like it has been divided exactly into 3 pieces. It is impossible to divide a physical object using mathematical precision.

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  • $\begingroup$ So, we can't have 3 identical objects? Because if we can have, the 3 objects together is 100% and when we take 1 it is exactly 1/3. or no? $\endgroup$
    – jm666
    Mar 30 '14 at 22:59
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    $\begingroup$ You missed this in the question: "Imagine an apple which was cloned two times, so the other 2 are completely equal in 'quantity'. " $\endgroup$
    – Vector
    Mar 31 '14 at 1:10
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    $\begingroup$ First you have to find an apple containing a multiple of 3 atoms... $\endgroup$
    – NoBugs
    Apr 2 '14 at 1:34
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    $\begingroup$ @NoBugs That's easy: probably one apple in three has that property. $\endgroup$ Apr 4 '14 at 7:37
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I guess it came out of confusion; Confusing the number 100 with the totality of something (100%, 1 unit) - However, the value 4 can be put into a unit and start being 100% of itself, you can divide 100% by 3 and get 1/3 and 0(»zero) left but you can't do it with the 4. -- I think, as @KeithS wrote I am making a switcheroo, although not intentional.

As writting this I thought that maybe you can't divide always 100% of something into 3 equal parts without any leftovers. What if given object has 4 atoms, how do you do it?

As of writting the above I thought, one object (apple with 4 atoms for an example) / 3 is different from saying 3 (4 atomed apple)equal objects / 3.

And so, 100% can't always be divided by 3 without any left.

Also, the edit that was made to my initial post was important to the extent that explained it may be a dumb question (as I'm not a mathematician nor am I in the field) it was a problem that I thought of and as it seems to be now just confusion.

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    $\begingroup$ There is nothing different between dividing $100\%$ by $3$ and dividing $1$ by $3$ without remainder. In some number systems (naturals, integers) you cannot. The number $\frac 13$ does not exist. In others (rationals, reals, finite fields with characteristic greater than $3$) you can. In "the real world of apples" you can divide with great but not perfect precision. $\endgroup$ Apr 1 '14 at 2:41
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Sigh. So many almost-good answers, so many actually-good comments.

From @hardmath's comment on the currently top-voted answer:

...one-third of $100\%$ is $33\frac{1}{3}\%$.

If we can divide apples into thirds, why not divide percentiles into thirds? This is a perfectly acceptable percentage.

If, however, you really wish to stick to decimal representations of fractions, then you are faced with the problem (implicit in the question) that $3$ does not evenly divide powers of $10$. This is because, since $10 = 2 \times 3$, $3$ and $10$ are relatively prime.

But this does not mean that there is no valid "pure-decimal" representation of $33\frac{1}{3}\%$. There are in fact two accepted ways to represent this percentage:

  • $0.33333...$ Here, the '...' means that the $3$'s are repeated indefinitely.
    • (If you are concerned by the idea of "infinite 3's" and think that this implies a contradiction or paradox, then consider that "the $3$'s are repeated indefinitely" is not quite the same thing as "there are an infinite number of $3$'s"; the difference is that the latter is an impossibility (given finite screen/paper space, time, etc), while the former is merely a definition that simply says "we understand '...' to mean that a fraction with a remainder that does not divide evenly into a power of $10$.")
  • $0.\overline{3}$ This is simply a different notation for $0.33333...$. I prefer it to the latter because it allows you to specify precisely which digits are repeated; for instance, the decimal expansion of $\frac{1}{47}$ is $0.\overline{0212765957446808510638297872340425531914893617}$, which is much clearer than 0.021276595744680851063829787234042553191489361702127659574468085106382978723404255319148936170212765957446808510638297872340425531914893617....
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  • $\begingroup$ But the result 0.3333... is what I didn't want. Because 3*0.3333... isn't 100%. In the accepted answer I can see in the graph how the totality of the circle can be divided in 3 without any left, by changing the base. $\endgroup$
    – RAO
    Apr 3 '14 at 8:21
  • $\begingroup$ You are not correct; 3*0.33333 is 1, which is 100%. The base is irrelevant; in my answer there isn't "any left," as you put it. $\endgroup$ Apr 3 '14 at 15:17
  • $\begingroup$ Kyle in your comment above there should be 0.33333... not 0.33333 The three continues dots are mandatory. Not to mention- When we say something like $0.33333... = \frac{1}{3}$ we actually mean that 0.3333.... is a number which is the limit of $0.3_{1}3_{2}3_{3}...3_{n}$ s.t. $n\to \infty$. The thing is adding more 3's makes it more close to $\frac{1}{3}$ so the limit is $\frac{1}{3}$ but it(the decimal expression 0.33333...) never becomes equal to $\frac{1}{3}$ $\endgroup$
    – user103816
    Apr 3 '14 at 16:42
  • $\begingroup$ @Anupam Sorry, typing on a smartphone; not sure how I missed the dots. I'm aware of the distinction. $\endgroup$ Apr 3 '14 at 16:51
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In reality you are approximating, but mathematics is more than reality so it prefer preciseness whenever possible.

Here $\frac{1}{3}\times100=33.333\cdots$ and $3\times 33.333\cdots=99.999\cdots$. I think this is what confused you.


But $100=99.999\cdots$

Let $$x={\Large99.999\cdots}\\\implies 10x=999.999\cdots=900+99.999\cdots=900+x\\\implies10x-x=900\\\implies9x=900\\\implies x=\frac{900}{9}={\Large100}.$$

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Here, in a practical way – I understand all your concepts, sirs – I was using a simple calculator on my computer, and looks like mine's got a 10 decimal digit precision. Which make this possible:

9 decimal "3s" 33.333333333 * 3 = 99.999999999

10 decimal "3s" 33.3333333333 * 3 = 100

If I divide 100 by 3, it'll give me 10 decimal "3s". And this expression was valid: 100/3 * 3 = 100

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    $\begingroup$ That's simply because your calculator is doing an approximation. $\endgroup$
    – Vector
    Mar 31 '14 at 1:12
  • $\begingroup$ No offense, but your answer is kind of dumb. As comeAndGo says, it's only an approximation. But in practice, approximations aren't actually bad. $\endgroup$ Apr 2 '14 at 18:28
  • $\begingroup$ Well, I thought the OP was trying to solve a problem instead of get to know all the concepts of divisibility. I should go back to stackoverflow, maybe :) $\endgroup$
    – FRAGA
    Apr 3 '14 at 21:50
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In base 10, no, not as long as you only allow finite digits. In base 3, we certainly can (or any other base divisible by three).

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The number 100 as the maximum percentage (unity) is nominal, just for people's convenience. We could have as well used 300% as one totality for three apples. So, we could divide it by 3 in base 10 and have 100% per apple.

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