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June 30

Mathematical tricks using numbers

I have to perform a show named "Mathemagic" in my school, therefore, I need some mathematical tricks. For example - choose any number and multiply it by 2, then add any number to the result and so on. I need some tricks like previous one. So, please help. Publisher54321 (talk) 11:21, 30 June 2013 (UTC)[reply]

Check out Category:Recreational mathematics. -- Jack of Oz ^[Talk] 12:11, 30 June 2013 (UTC)[reply]

Thank you JackofOz. It would be more useful if someone suggest some videoes like this. Publisher54321 (talk) 13:27, 30 June 2013 (UTC)[reply]

Then your best bet is probably to Google around – there will be many examples around, but you'll have to select from amongst them what would suit your purposes. I suspect that the people who would respond here are no more likely than anyone else to know interesting tricks of this nature off the tops of their heads. — Quondum 14:34, 30 June 2013 (UTC)[reply]

You might be able to demonstrate the birthday problem or the monty hall problem. SemanticMantis (talk) 17:57, 30 June 2013 (UTC)[reply]

Google "proof that 1 equals 0" and "proof that 2 equals 1" for amusing, simple errorneous proofs that demonstrate mathematical fallacies — or you can even check out that specific article on fallacies for more. —SeekingAnswers (reply) 20:27, 30 June 2013 (UTC)[reply]

[1]. Count Iblis (talk) 20:34, 30 June 2013 (UTC)[reply]

Learning to do a perfect Faro shuffle, and performing it 8 times in a row will restore the original order of the deck. Use this to explain modulo arithmetic. For more ideas, this ted talk will probably be helpful. MChesterMC (talk) 08:37, 1 July 2013 (UTC)[reply]

1/9801 = 0.00010203040506... and so on, the decimal representation contains every two digit number in order, except 98. Similar fractions exist, 1/81 is single digit numbers excluding 8, 1/998001 is three digit numbers excluding 998, etc. Aside - is there a Wikipedia article on this? It's a curious bit of math but perhaps not notable. Organics^LRO 09:11, 1 July 2013 (UTC)[reply]

These are simply numbers of the form (10ⁿ-1)². There is no Wikipedia article, but the numbers are discussed as OEIS sequence A059988. —SeekingAnswers (reply) 13:47, 1 July 2013 (UTC)[reply]

Does anyone have a nice explaination as to why this happens? (feel free to move this into its own section if you think that would be more appropriate) MChesterMC (talk) 16:01, 1 July 2013 (UTC)[reply]

A slightly handwaving proof is to use long division. First note that 1/9 = .11.., 1/99 = .0101.., 1/999 = 0.001001... If you take the third of these as an example and do long division by 999 you will see how the sequence of 001, 002, 003, ... comes about, and with a bit of thought why one of the numbers is missed out. AndrewWTaylor (talk) 17:50, 1 July 2013 (UTC)[reply]

Copeland-Erdős constant and normality

I was reading the normal number article, which states that Arthur Herbert Copeland and Paul Erdős proved in 1946 that the Copeland-Erdős constant is normal. This result surprises me, because since all prime numbers other than 2 are odd, all prime numbers other than 2 end in an odd digit, so I would expect skew of the digit distribution toward odds, since each prime number other than 2 is guaranteed at least 1 odd digit, while there is no such at-least-1-digit guarantee for even digits. So, my questions:

(1) Thinking about it some more, my guess at what is happening is that as the prime numbers go toward infinity, prime numbers become so long digit-wise that the oddness of the last digit becomes negligible. Is this in fact the explanation?

(2) Looking at the first few digits (0.235711131719232931374143...), it's obvious that odds far outnumber evens within the early digits. But since the constant is normal, the evens must "catch up" eventually: either...

(a) ...the evens asymptomatically approach from below a 50% distribution of all digits, or...
(b) ...(what seems to me much more likely) which parity of digits are ahead changes infinitely often, though it might take a long time and a very large prime for the evens to first catch up (reminding me of the very large Skewes' numbers and related numbers wherein π(x) finally catches up to li(x) for the first time).

(Well, okay, I suppose there's a third possibility, a combination of the above two cases so that after a finite number of switches of the lead, one parity stays ahead while the other stays asymptomatically close, but I don't think that's likely.) Does anyone know if there is a proof of which case is true? If, as I suspect, case (b) is true, what is the smallest prime at which the cumulative even digit distribution catches up to the odds?

—SeekingAnswers (reply) 20:08, 30 June 2013 (UTC)[reply]

(1) Yes, a single digit in each prime doesn't matter. That's also why it doesn't matter that no prime starts with 0. In fact, a fixed number of digits in each prime wouldn't matter. If we inserted a million 1's or a googolplex 3's between each prime then the resulting number would still be normal. PrimeHunter (talk) 23:22, 30 June 2013 (UTC)[reply]

Here's how I would approach it: For a given n, there number of n+1-digit primes is approximately

{\frac {9\times 10^{n}}{n\log 10}}

(see prime number theorem; we could get better estimates but this will probably be good enough). For each of these primes, the last digit is even, and the others may be assumed to be essentially "random" (this is a heuristic technique rather than rigorous math, but it usually works in practice). So we can take the pool of digits added because of these primes to be even with a binomial distribution with

p={\frac {{\frac {n}{2}}+1}{n+1}}\approx 1+{\dfrac {1}{n}}

and sample size

{\frac {9\times 10^{n}}{n\log 10}}

. Now try to get a sense of the chance that there are more odd digits than even digits among n+1-digit primes, and try to figure out how often that would have to happen for the odd digits to catch up. --Trovatore (talk) 23:23, 30 June 2013 (UTC)[reply]

I think you wrote "even" several times (for example, "for each of these primes, the last digit is even" and "for the odd digits to catch up") when you meant "odd", and vice versa. —SeekingAnswers (reply) 00:19, 1 July 2013 (UTC)[reply]

Whoops, good catch. Doesn't change anything, though. --Trovatore (talk) 00:25, 1 July 2013 (UTC)[reply]

Thanks. It's a nice, sensible heuristic method, though unfortunately, that kind of argument can only tell us rough bounds for the approximate size under which evens are likely to catch up, without giving any exact answer. It could also be the case that due to a run of primes with an unusual distribution, the evens first catch up much earlier or much later than the expected size. Anyone know if there has been any methods or results published on this problem that are more specific/exact? —SeekingAnswers (reply) 01:04, 1 July 2013 (UTC)[reply]

I was hoping you'd work through it a bit more. I'm almost sure that the heuristic's answer will be, the evens will (with very high probability) never catch up, because the chance of there being more evens than odds in the block of n+1-digit primes will fall off so quickly that the infinite product will converge. --Trovatore (talk) 01:09, 1 July 2013 (UTC)[reply]

You know, I went for a walk, and it occurred to me that there's a complication. Every time you go to a new number of digits (actually, every time the first digit changes), for a while, all the primes are going to have a long string of zeroes starting at the second digit. I haven't written anything down yet, but if I did it right in my head, I now think that probably that's eventually going to make the evens catch up for a little while after every change of the first digit. --Trovatore (talk) 04:37, 1 July 2013 (UTC)[reply]

Those zeroes would have to overcome the additional lead in odds gained by a long string of nines starting at the second digit that occurs before we get to the new first digit, though. —SeekingAnswers (reply) 20:19, 4 July 2013 (UTC)[reply]

Okay, I got some experimental results now.

Let r(n) be the proportion of even digits after the nth prime. So, since the constant starts out 0.2 3 5 7 11 13 ..., the first few values of r(n) are r(1) = 100%, r(2) = 50%, r(3) = 33.333...%, r(4) = 0.25%, r(5) = 16.666...%, r(6) = 12.5%. Below, when I refer to the "maximum value" of r(n), I am disregarding the trivial r(1) and r(2) values.

I prepared an Excel spreadsheet to calculate values of r(n). Then I wrote a Perl script to calculate r(n) more efficiently. As a check against formula or programming errors, I compared the results of the two against each other for the first 100,000 primes, verifying that the calculated results were identical for the 2 very different methods. Then I ran the Perl script up to n = 7.5 x 10⁷ (75 million); if you're wondering how large these primes are, the 75,000,000th prime is 1,505,776,939.

For n ≥ 3, r(n) initially falls before starting to rise, before finally tying r(3) = 1/3 at r(380), with r(381) = 444 / (444 + 883) ≈ 33.45% being the first value of r(n) to exceed r(3).

Beyond r(381), r(n) oscillates (obviously), but on average, it rises much more than it falls and initially grows rapidly on average — but as the primes get larger and larger, its average rate of growth falls. r(n) first hits 34% at r(389), hits 35% at r(416), hits 36% at r(654), hits 37% at r(1,106), hits 38% at r(3,097), hits 39% at r(6,861), hits 40% at r(24,613), hits 41% at r(55,426), hits 42% at r(210,117), hits 43% at r(1,790,106), and hits 44% at r(25,609,981).

Anyway, as of the 75th million prime 1,505,776,939, the highest value of r(n) thus far is 44.2537565841856...% at the 46,450,161st prime, 909,090,109. So, yeah, I still don't know if r(n) does ever hit 50%. It might, or it might not. I still think that it will, but if it does, it clearly is going to take a long time and a very large prime to do so.

Looks like this is still an open question, so anyone else's further input or ideas are welcome. :)

—SeekingAnswers (reply) 07:59, 3 July 2013 (UTC)[reply]

A. Numerical solutions to 4/(4n+1) B. The problem 5/(5n+1)=1/a+1/b+1/c

A. Where can I find numerical solutions to the Erdos-Straus conjecture for numbers that remain after Mordell's results (i.e., for 1, 121, 169, 289, 361, and 529 modulo 840). I am especially interested in the numerical solutions for 169, 361, 1009, and 1201.

B. Wikipedia of course has an article on the Erdos-Straus conjecture. But is there also an article on - or even a name for - the similar problem 5/(5n+1)=1/a+1/b+1/c? — Preceding unsigned comment added by BH1221 (talk • contribs) 20:11, 30 June 2013 (UTC)[reply]

I'm pretty sure there's no specific article for the 5-numerator case. Near the end of the Erdos-Straus conjecture article, the 5-numerator case is mentioned in passing, though. Quoting from the article:

A generalized version of the conjecture states that, for any positive k there exists a number N such that, for all n ≥ N, there exists a solution in positive integers to k/n = 1/x + 1/y + 1/z. The version of this conjecture for k = 5 was made by Wacław Sierpiński, and the full conjecture is due to Andrzej Schinzel.

Furthermore, in OEIS sequence A073101, which is the number of solutions to the 4-numerator case, the comments refer to "Sierpinski's conjecture for 5/n", so that is the name for your related problem.

That OEIS sequence write-up also mentions that "solutions can be printed by removing the comment symbols from the Mathematica program" that is given there, so that will allow you to find the numerical solutions for the specific cases you mentioned.

—SeekingAnswers (reply) 20:20, 30 June 2013 (UTC)[reply]

Is Mathematica now free? The last time I looked it cost a few hundred dollars.

(Addition: In Part A of my original question, I forget to say that I was interested also in the numerical solutions for n=193 and n=1033.) — Preceding unsigned comment added by 82.81.156.242 (talk) 10:40, 1 July 2013 (UTC)[reply]

Mathematica is not free. But if you're a student at a university, you should be able to use it through your university's institutional license. Even some high/secondary schools will have institutional licenses. If you still don't have access to Mathematica, you can try to adapt the code for some free computer algebra system like PARI/GP or Sage, or some general-purpose programming language. —SeekingAnswers (reply) 20:01, 1 July 2013 (UTC)[reply]

I'm not a student so Mathematica is out, and (according to Wikipedia) Sage is not compatible with Windows (my computer's system). Perhaps I'll try adapting the code to that old standby, Fortran. — Preceding unsigned comment added by 82.81.156.242 (talk) 21:12, 1 July 2013 (UTC)[reply]