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BS: Who else plays little # games?
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Subject: BS: Who else plays little # games? From: MGM·Lion Date: 30 Aug 13 - 02:44 AM Although entirely unmathematical, (a great regret to me but there it is), I like to play little games with numbers. I enjoy reading of the number games and discoveries by Charles Dodgson [aka Lewis Carroll] recorded in such places as Martin Gardner's "The Annotated Alice" - but, then, Dodgson was a distinguished Oxford mathematician. E.g. it occurred to me yesterday that my present age, 81, is a nice number, because it is a square. Moreover, its digits add up to the number that it is the square of. Do any other squares do that? And does anyone else find such things of as much interest and entertainment as I do? Or have any examples to contribute? ~Michael~ |
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Subject: RE: BS: Who else plays little # games? From: Bee-dubya-ell Date: 30 Aug 13 - 06:14 AM 81 squared is 6561. 6+5+6+1=18. Rotate your computer screen 180° and... 81! |
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Subject: RE: BS: Who else plays little # games? From: MGM·Lion Date: 30 Aug 13 - 06:31 AM Thank you BWL. This is now getting a bit out of my league: but I note that the number you produce is the square of 'the number I first thought of', as the catch-problems used to say.. Is that why the phenomenon of the added digits repeats. And will the same thing happen by adding the digits of 6561²? No -- their sum appeared to be 27: the cube of the square root of 18/2 Now, where does that fit into the relationship? See what I mean about not being any sort of mathematician. One of those would presumably have extrapolated some general principle from all this. But not me tho... LoL |
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Subject: RE: BS: Who else plays little # games? From: Nigel Parsons Date: 30 Aug 13 - 08:28 AM Any multiple of 9 will have digits which also add up to a multiple of 9. And will the same thing happen by adding the digits of 6561²? No -- their sum appeared to be 27 and the digits of 27, 2+7=9 This is an extension of the fact that any multiple of 3 has digits which add up to a multiple of 3. |
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Subject: RE: BS: Who else plays little # games? From: DMcG Date: 30 Aug 13 - 12:24 PM There's a whole bunch of 'magic' tricks that rely on that property of nine (or, for the really obscure ones, the digit N in numbers of base (N+1)) In fact, if I remember correctly, Martin Gardner had a chapter on that in his book 'Mathematical Puzzles and Diversions' (Or perhaps his other book 'More Mathematical ...') One concerned a sort of telephone dial with weird symbols, but they were just flummery to hide the numerical basics. |
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Subject: RE: BS: Who else plays little # games? From: DMcG Date: 30 Aug 13 - 12:28 PM And one thing I pointed out to my daughter recently was that, short of a major improvement in life maintenance, she is just about to pass through the last of the only three ages with a specific property: 1-to-the-power-1 => 1 2-to-the-power-2 => 4 3-to-the-power-3 => 27 4-to-the-power-4 => ... unlikely! |
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Subject: RE: BS: Who else plays little # games? From: TheSnail Date: 30 Aug 13 - 07:55 PM I raher like this (but it's easier to check it with a calculator) - 1/7 = 0.142857* 2/7 = 0.285714* 3/7 = 0.428571* 4/7 = 0.571428* 5/7 = 0.714285* 6/7 = 0.857142* where * means recurring. The six digits repeat forever. |
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Subject: RE: BS: Who else plays little # games? From: Rapparee Date: 30 Aug 13 - 09:41 PM My current license plate is 1B S5762. In 1957 I was starting seventh grade. In 1962 I was starting my last year of high school. The 1BS is self-explanatory and, according to some people, very appropriate. |
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Subject: RE: BS: Who else plays little # games? From: Airymouse Date: 30 Aug 13 - 11:09 PM Snail's 1/7 2/7 etc. is deeper than it looks: After you know the first 3 digits 142 you get the next three digits by subtracting from 9 (e,g. 8=9-1, 5=9-4, etc. and multiplication merely rotates the digits unless you multiply by a multiple of 7. (142857)(3) = 428571. If you multiply by a number larger than 7 the number still rotates, but you have to help by joining a few digits by adding e.g., (142857)(23)=3285711 and you have to add the first and last digits to see the rotation. What makes this work? The key is that when you divide 7 into 1 by long division it repeats after 7-1 digits, because you get all the possible nonzero remainders (1,2,3,4,5 and 6). Are there other such numbers (like 7, which yields .142857...)? Yes and they are all prime numbers The next one is 17 1/17 repeats after 16 digits. Most pocket calculators show only the first 8 digits 0588235 but 9-0 =9 etc so the full 16 digits are .0588235294117647. Are there infinitely many such numbers? This is an unsolved problem. The statement is true if Artin's conjecture is true. |
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Subject: RE: BS: Who else plays little # games? From: Rapparee Date: 31 Aug 13 - 12:09 AM License plates are something that drives me nuts. For example, I'll see one with a number like, oh, 4679. Right off I think, "That's a neat number becase 4 and 6 are even and 7 and 9 are odd and they come in a sort of sequence. Or when I got to choose my cell phone number I chose 1861 because that was the year the US Civil War started and I'd remember it. My wife got 4903 by the luck of the draw, and that's just two numbers off from our postal box number. Or I'll mentally add all of the numbers together (3496 might get 3+4+9+6=22=2+2=4 and square root of 4 is 2 but the cube root of 4 (first number is 3) is about 1.6 when I do it by multiplying numbers until I get the closest to 4 (it's really about 1.5874010519681994 -- I looked it up). Or to go to sleep I'll mentally calculate the decimal answer to a problem like 35/17, which fractionally is 2 1/17th, but I have to arithmetically calculate the decimal equivalent of the fraction (1/17 x 100/1 = 100/17 = about slightly less than 6. Usually by then I'm asleep, dreaming of Fourier transformations or something. Try it yourself -- mental math can put you right to sleep, just as it did in grammar school. |
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Subject: RE: BS: Who else plays little # games? From: MGM·Lion Date: 31 Aug 13 - 02:40 AM I have always liked, as wikipedia tells it, the 'famous anecdote of the British mathematician G. H. Hardy regarding a visit to the hospital to see the Indian mathematician Srinivasa Ramanujan [1887-1920]. In Hardy's words: " I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."' What are the two ways? Work them out. Even I did. ~M~ |
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Subject: RE: BS: Who else plays little # games? From: TheSnail Date: 31 Aug 13 - 04:33 AM Thanks for that Airymouse. Adds things I never got round to finding out epecially that the next number is 17. Your method of generating the second three digits from the first three has interesting consequences. 1/7 + 6/7 = 7/7 = 1 but try adding 0.142857* and 0.857142*. I'm a bit lost by your bit about multiplying by a number larger than 7. Surely, for instance, 8 * 1/7 is 1 + 1/7 so the decimal part behaves just as before but the integer part can be anything. |
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Subject: RE: BS: Who else plays little # games? From: DMcG Date: 31 Aug 13 - 04:41 AM Ah, but, O Molluscy one, 1 is exactly the same number, mathematically speaking, as 0.9999999999 recurring. I know it seems odd to 'outsiders', but that's how it is. It's to do with the mathematical definition of equality. Now, I know it might seem that equality is easier to define than 'less than', but it happens that to mathematicians 'less than' is the more important relationship. And equality on numbers is defined like this. Two numbers are equal if for any number, epsilon, that you care to pick, however small, the difference in the two numbers is less than epsilon. |
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Subject: RE: BS: Who else plays little # games? From: DMcG Date: 31 Aug 13 - 05:03 AM That's what comes of trying to translate maths into English. When I say 'for any number, epsilon', read 'for every number, epsilon'. But since I've been obliged to repost, I'll take the chance to explain a little more. '1', '0.999 recurring', etc, are not, strictly speaking, numbers, but representations of numbers. Every (rational) number has lots of different representations: In fractional notation we could have 1/2, 2/4, 3/6 and so on, all representing the same number. We are used to that, so we are taught in schools to always use the simplest form, but all the others are equally valid. And it is the same in decimal notation: every rational number has two representations, and we always pick the simplest. So we normally write 0.5 rather than 0.499999recurring, but both represent exactly the same number in the way 1/2, 2/4 do. |
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Subject: RE: BS: Who else plays little # games? From: TheSnail Date: 31 Aug 13 - 05:26 AM DMcG 1 is exactly the same number, mathematically speaking, as 0.9999999999 recurring. I know, but in the context of this thread which has a range of countributors from MtheGM who describes himself as "entirely unmathematical" to people such as Airymouse and your good self who are clearly highly mathematiccally educated (I'm somewhere inbetween) I think it has a certain "Well I never" factor. I have seen a proof of 1 = 0.999 recurring but can't remember it. Do you have it to hand? |
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Subject: RE: BS: Who else plays little # games? From: DMcG Date: 31 Aug 13 - 05:41 AM The proof follows from the definition I gave for equality, where we look at the difference between the two numbers. In this case we want the difference between 1 and 0.9999recurring. Now, that recurring is a bit tricky, so let's take a series of finite approximations 1-.9 = 0.1 1-.99 = 0.01 1-.999 = 0.001 1-.9999 = 0.0001 You can see where this is going, I hope. For any epsilon you care to pick, let's say, 0.00000000000000000000000000000000001, or one with a hundred million leading zeros before the '1', I can produce a term in this series where the difference is smaller, because the 'recurring' means I can write as many nines as I like, even if it is a hundred million and one. And since that's my definition of equality, the two representations denote the same number. |
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Subject: RE: BS: Who else plays little # games? From: TheSnail Date: 31 Aug 13 - 05:54 AM OK, that's a proof but not quite what I had in mind. Perhaps I should have said something like demonstration. Found it! 1/9 = 0.1111.... 9 * 1/9 = 9 * 0.1111.... 1 = 0.9999.... |
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Subject: RE: BS: Who else plays little # games? From: catspaw49 Date: 31 Aug 13 - 09:52 AM When I taught school I did a math segment in AutoTech to my usually non interested teen males. Most were very poor in math and had learned to hate it. I started out with number games and especially a few nine games which always fascinated them, caught their interest, and at least made them more easily taught. You have to be more entertaining than MTV.......... Spaw |
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Subject: RE: BS: Who else plays little # games? From: Rapparee Date: 31 Aug 13 - 10:26 AM Spaw, teaching (from a transcript): Okay, you dipshits. Lissen up. You take the number 9, see? Only make it with a straight tail. Flop it over and you got either a 6 or a lower-case B -- that's a little B to you shitheads. Lay is down on the round part and it looks like that thing you see in your camera, right? Lay it down on the flat part and it looks like Whatsisname over there in the corner, only without the peter to play with like he's doin' now. Learn this, because it's important. Why is it so fuckin' important? Because I said so, you little.... It goes on like that for every single number up to a googleplex. |
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Subject: RE: BS: Who else plays little # games? From: Airymouse Date: 31 Aug 13 - 10:41 AM Snail. Sorry about my muddling. I meant 8 times 142857, the product to two whole numbers. The product is 1 14285 6. I isolated the first and last digits of this product, because they are the digits you have to add to get "142857". For some psychological reason that I don't understand, people seem happy with 1/3= .333333... and 2/3= .6666666... but when you add 1/3 + 2/3 they are puzzled by 1=.9999.... Of course infinite decimals really are something different from ordinary arithmetic. You can add only two numbers at a time, so if Lewis Carroll asks what's 1 and 1 and 1 and 1, Alice should be asking herself is this ((1+1)+1) +1) or is it (1+ (1+1)) +1 or is it .... But we all know that no matter which of the possibilities Lewis has in mind the answer is 4. Putting this quibble aside finite decimals make sense .385 means 3/10 + 8/100 + 5/1000. Since you can add only two numbers at a time, no matter what order you choose you will never be able to add up the infinite list 9/10 + 9/100 + 9/1000 etc. Thus, infinite decimals are a different animal than finite decimals, which are just the finite sum of fractions. |
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Subject: RE: BS: Who else plays little # games? From: GUEST,Guest from Sanity Date: 31 Aug 13 - 11:04 AM Rapparee: "License plates are something that drives me nuts. For example, I'll see one with a number like, oh, 4679. Right off I think..." A few years back my wife and I had license plates on each of our cars....Mine: OI12EU ..Hers: OI12EU2. People who 'got it' thought it was pretty cool! Regards, Rap.. GfS |
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Subject: RE: BS: Who else plays little # games? From: Mr Happy Date: 31 Aug 13 - 11:59 AM .............but why do I always remember x= -b +/- sq.root.b squared - 4ac 2a ?? When I've never ever had to use it in everyday life?? |
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Subject: RE: BS: Who else plays little # games? From: GUEST Date: 31 Aug 13 - 02:24 PM I can't answer that one, Mr Happy, beyond saying it is amazing how much mental clutter we pick up along the way. Airymouses' comment is more subtle. It took mathematicians literally centuries to get a firm grip on infinite series and thinking of a decimal representation of a number as such a series is valid. But it is a mistake to think of the mechanical process of adding digits and hence conclude recurring decimal numbers are in some way in a different class to those we can add up in a finite number of stepps |
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Subject: RE: BS: Who else plays little # games? From: GUEST,DMcG Date: 31 Aug 13 - 03:31 PM Sorry, that was me above and I got cut off mid sentence. The mistake is getting too hung up on the method of getting the answer rather than the answer itself. That way, instead of concluding that the terminating and recurring numbers are in some sense distinct, you start wondering if the problem lies in the method of adding the digits rather than the numbers themselves. Obviously enough, mathematicians don't handle infinite series by adding all the terms together explicitly since by definition that is an infinite process. Exactly how the do it varies but a common way is to prove the sum of the series must be X or more and simultaneously X or less. Hence you can work out the sum without actually doing the calculation. And to get back onto representation again: there's no need to stick to a base-10 notation. If you use a base 7, for example, a number like 2/7 that is recurring in base 10 'decimals' is not in base 7 'septimals'. So a suggestion there are distinct types of number depending on whether they recur of not is clearly an artifact of the chosen notation, not of the number |
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Subject: RE: BS: Who else plays little # games? From: TheSnail Date: 01 Sep 13 - 04:20 AM With you now Airymouse. I hadn't realised you were dealing with the six digits in isolation. If you were still looking at the recurring decimal, the 1 you add to the 6 in 1 14285 6 would be the carry from the next set of recurring digits 1.142856 + 0.000001142856 + 0.000000000001142856 ... = 1.142857142857142857... |
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Subject: RE: BS: Who else plays little # games? From: Mysha Date: 02 Sep 13 - 12:37 AM Hi, Well, on the one hand: 1*1= 1; adding all the digits of 1 gets you 1. Yet, on the other hand: What if I choose (1 - 0.9*)/2 for epsilon? Bye Mysha |
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Subject: RE: BS: Who else plays little # games? From: Rapparee Date: 02 Sep 13 - 12:18 PM Mysha, you'll softly and silently vanish away. Another thing to mess your mind up with is: If 1+1=2, then the sum of any number added to itself is the next number in the sequence: 2+2=3, 3+3=4, etc. This would mean that 1+2=1.5. Try not to get your pay calculated this way. |
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Subject: RE: BS: Who else plays little # games? From: GUEST,DMcG Date: 02 Sep 13 - 02:18 PM That was why, Mysha, I had to change the phrase 'any epsilon' (which is ambigous in English, to 'every epsilon'. Having a single epsilon can't on its own, show equality (though a single could disprove equality) I don't go on like this in real life, honest! |
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Subject: RE: BS: Who else plays little # games? From: GUEST,Grishka Date: 03 Sep 13 - 07:33 AM I found Mysha's post quite to the point. Her or his first assertion gives one more solution to the OP's question, and the other one points to some sloppiness in DMcG's writing. Hypothesis: (1 - 0.9*) is smaller than any arbitrary epsilon. Mysha's "disproof": (1 - 0.9*)/2 for epsilon is even smaller. Pedants, however temporary, should take particular care to be exact and correct themselves. |
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Subject: RE: BS: Who else plays little # games? From: GUEST Date: 03 Sep 13 - 08:54 AM Not a disproof or a revelation of my sloppiness either, Grisha, though I admit to quite a bit of sloppiness in my postings in general, and indeed put in a correction to the definition of equality on this thread precisely because of a sloppiness on my first attempt. It is difficult to avoid pendantry in maths precisely because being exact is important, so I do plead guilty with mitigating circumstances on that one. Rapparee pointed out with some subtlety why Misha's disproof wasn't - think about his post a little more |
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Subject: RE: BS: Who else plays little # games? From: GUEST,DMcG Date: 03 Sep 13 - 09:09 AM I should add to that, of course, that any attempt to translate maths into ordinary English risks ambiguity or omission. I did not say, though it was my intention, that epsilon should be bigger than zero, as that is a well understood convention is science generally. Clearly, an epsilon of -1000 would make a lot of numbers equal that we did not intend. If you want to include that as an example of sloppiness, feel free. While we are at it, I also assumed scalar quantities, not, for example, complex numbers. All such things would have been explicit had I used formal notation not English, but it would have limited the readership even more than tedious pedantry (and is painful to enter as html) |
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Subject: RE: BS: Who else plays little # games? From: MGM·Lion Date: 03 Sep 13 - 09:58 AM This has now turned into BIG number games. The numerate have hijacked the thread, dammit, leaving all us poor old games-playing hammer-chooers out in the cold! Chiz-chiz! ~M~ |
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Subject: RE: BS: Who else plays little # games? From: Keith A of Hertford Date: 03 Sep 13 - 10:28 AM Only our generation Mr Happy. The formula is now provided in GCSE papers. |
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Subject: RE: BS: Who else plays little # games? From: GUEST,DMcG Date: 03 Sep 13 - 11:48 AM Oh, I hope not, MtheGM. There's a good case for all sorts of number games just for the fun of it, and many a deep theory is inspired by a simple game, exactly is it is in science or, analoguously, literature. Lear has his place, not just Tolstoy. Not everything needs the PhD, whether maths, english or music. "There is beauty in the bellow of the blast, There is grandeur in the growling of the gale ..." |
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Subject: RE: BS: Who else plays little # games? From: GUEST,Grishka Date: 03 Sep 13 - 12:30 PM DMcG, sometimes I enjoy playing the temporary pedant myself, but not this time. Mysha (who is male, as my private directory of Mudcatters tells me) did not provide a disproof - how could he - but a "disproof", i.e. as precise as your "proof", and therefore pointing at that lack of precision. No need though to mend it, though. It is basically high school maths, not requiring higher academic studies. Still, regarding Airymouse of 31 Aug 13 - 10:41 AM, it is noteworthy (though not "puzzling") that all numbers that have a finite decimal representation also have another, infinite one (ending with an infinitely many 9s), whereas all other numbers only have one representation. In contrast, MtheGM's original question, which properly reads "Are there infinitely many numbers with that property, and if not, which ones or how many or which is the largest one?", does not look easy to me at all. A friend of mine who minored in maths gave up after ten minutes. I would not bet that every PhD mathematician performs better. |
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Subject: RE: BS: Who else plays little # games? From: gnu Date: 03 Sep 13 - 12:38 PM Laff and learn. This guy is gooood. |
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Subject: RE: BS: Who else plays little # games? From: DMcG Date: 03 Sep 13 - 01:25 PM Grisha - I reckon we should leave this one, as we risk completely alienating the MtheGM's of this world. I'll just say I used the term "disproof" not be way of an accurate definition, but because that's how it had been labelled earlier. I've reread what I wrote and stand by it as long as we include the caveat I gave in my 03 Sep 13 - 09:09 post (epsilon bigger than zero). |
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Subject: RE: BS: Who else plays little # games? From: TheSnail Date: 03 Sep 13 - 02:28 PM I wouldn't mind a bit of feedback on my contribution of 31 Aug 13 - 05:54 AM. Does that work as a proof and if not, why not. No axe to grind; I'm here to learn. |
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Subject: RE: BS: Who else plays little # games? From: DMcG Date: 03 Sep 13 - 02:50 PM It works as a demonstration fine. The problem from the point of view of pure mathematics is that it is an infinite process. Now, when you represent a number like, oh, 0.7687326 to one less decimal place the convention normally used is that you round the penultimate digit up if it is 5 or more, and down if it less than five (at least, that's the normal convention - there are weirdnesses like banker's rounding, but we will ignore that!), this would be0.768733. Mow, if we have 0.99999999... to any *finite* number of places, and you wish to stop writing the places out, then the rounding rules take you to 1.00000000.... which does equal 1, but not in an interesting way. Conversely, if you use the other common convention and truncate, you have 1=0.9999 (say), which isn't true. Either way, you don't really fully express the concept. And if you want to 'go on forever' you are on treacherous ground: infinite series are vicious beasts. By way of illustration, and for a spot of light-hearted maths (!!!!), think about the series S = 1-1+1-1+1... What does it add up to? Well, lets bracket things this way S = (1-1) + (1-1) + (1-1) ... ===> S = 0 But noting that a-b is the same as a+(-b), a different bracketing gives S = 1 + (-1+1) + (-1+1) + (-1+1) ... ===> S = 1 Or even S = 1 - (1-1+1-1+1-1..) ... ===> S = 1 - S ===> S = 0.5 It's problems like that which made mathematicians very wary of infinite series and the need to treat 'em with kid gloves. |
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Subject: RE: BS: Who else plays little # games? From: GUEST,Grishka Date: 03 Sep 13 - 03:22 PM DMcG and TheSnail, I do not think it is of interest to other readers. (If you are interested privately, a hint, courtesy of my high school teacher: the epsilon thing is about sequences or series. See Wikipedia, also at Decimal.) |
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Subject: RE: BS: Who else plays little # games? From: TheSnail Date: 03 Sep 13 - 05:20 PM Grishka, I would have thought that my method was more accessible to other readers than the epsilon stuff. I reproduce it here - 1/9 = 0.1111.... 9 * 1/9 = 9 * 0.1111.... 1 = 0.9999.... I may have been a bit sloppy with the notation for recurring. It could have been - 1/9 = 0.(1) 9 * 1/9 = 9 * 0.(1) 1 = 0.(9) but I think people knew what I meant. I'm still unclear as to what is specfically dodgy about it. |
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Subject: RE: BS: Who else plays little # games? From: GUEST Date: 03 Sep 13 - 05:36 PM Not dodgy; just to make a formal mathematical proof out of it you have to demonstrate that the power series that the notation is shorthand for is well behaved and that would take a few more lines and need a few more powerful tools. But nothing fundamentally wrong, no. |
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Subject: RE: BS: Who else plays little # games? From: MGM·Lion Date: 03 Sep 13 - 05:44 PM DMcG -- Indeed. But I can keep up with Tolstoy as well as Lear -- whereas... Ah well... ~M~ |
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Subject: RE: BS: Who else plays little # games? From: GUEST,Grishka Date: 03 Sep 13 - 06:29 PM That anonymous guest was not I, but I agree. |
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Subject: RE: BS: Who else plays little # games? From: GUEST,Grishka Date: 04 Sep 13 - 07:13 AM My friend has taken the challenge of the OP and asserts that 1 and 81 are the only positive integers with that property. (Admitting 0 makes for an additional solution.) His proof (sketched): assume the number to have n decimal places, and x being its square root (both positive integers), then 10n ≤ x². Since x is the sum of those decimal digits, x ≤ 9n. Everything being positive, we can square this to get x² ≤ 81n². Together, we get 10n ≤ 81n². This is not true for n=3 (1000 being greater than 729), and it cannot be true for any higher n, roughly because the left side grows exponentially. ("The details are left to the reader.") Thus n can only be either 1 or 2, easily seen to yield no more than the two solutions mentioned. All school maths, nothing esoteric. |
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Subject: RE: BS: Who else plays little # games? From: MGM·Lion Date: 04 Sep 13 - 07:33 AM _All school maths, nothing esoteric. _ Oh goody. How consoling... |
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Subject: RE: BS: Who else plays little # games? From: TheSnail Date: 04 Sep 13 - 09:17 AM Anonymous GUEST (is that you, DMcG?) But nothing fundamentally wrong, no. O, good. I think it fits well with little # games His proof (sketched): assume the number to have n decimal places, and x being its square root (both positive integers), then 10n ≤ x². I've got a bit more than school maths but you've lost me already. |
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Subject: RE: BS: Who else plays little # games? From: DMcG Date: 04 Sep 13 - 10:55 AM Yes, it was I. My phone doesn't like holding cookies so I often forget to check whether it still remembers who I am. Going back to the original question: I haven't spent the ten minutes investigating this because it is a much harder problem than it first appears. The difficulty is that it is a mixture of a properties about numbers and properties about notation. In particular when you get a carry' it becomes really difficult to express 'the sum of the digits'. On the other hand, the 81 is part of a more obsure set that is quite interesting: suppose you write a number in base N rather than base 10. Then, for every N there is a number that has the same property of the sum of the digits matching the square root. So for example 41 in base 6 is 25 in base ten, which is the square of 4+1. This doesn't need more than school algebra to work out what the number is, but even so it might take a few minutes. The Eureka! Moment when you see how to solve it might make the exercise worth it. |
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Subject: RE: BS: Who else plays little # games? From: GUEST,Grishka Date: 04 Sep 13 - 12:58 PM Snail, you are right, it must be n-1 (all my own error when reformulating and trying to simplify). You can fix the proof though, I am sure, and the assertion stays the same. |
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Subject: RE: BS: Who else plays little # games? From: TheSnail Date: 05 Sep 13 - 06:02 AM You mean "10n-1 ≤ x²"? Fine, but now if n=3... |
