The Millionth digit.
 

:smile:

If you write the sequence of numbers one after the other in their natural order without separating them, as though one were writing a single number:
123456789101112131415...

Find what digit occupies a given position, say the millionth ?

I would prefer a non computer solution.

Mally :coffee:

A part of problem 40 at [url]http://projecteuler.net/index.php[/url]. There are 9*10^(x-1) of x-digital numbers (x>=1).First find n for which sum of length of all numbers f(n) consists of no more then n digits, less and closest to 10^6. f(n)=sum(9*x*10^(x-1),i=1..n)=1/9*(1-10^n+9*n*10^n). For n=5, f(n)=488889 -the closest value less then 10^6. Then find the ammount of 6-digits numbers, which sum of length less and closest to 10^6-f(5)=511111,The number is [511111/6]=85185. Thus, we have that all digits of all numbers in range (1..185184 (cause the 100000 is the first 6-digits number)) are included in our sequence (no more then 10^6 digits). Number of digits of 185185 which lie in sequence is equal 511111-85185*6=1. Thus, the millionth digit of sequence is the first digit of 185185- which equals 1.

This problem can be solved in common case, but I suppose that symbolic result would be quite huge to be evaluated and posted there.

Here's a kinda neat way to get the solution using a similar type of sequence:

The millionth digit is given by the
(1000000 - (9 * 54321))%6 th digit of the
(1000000 - (9 * 54321))/6 th 6 digit number (which has already been pointed out to be 185185) or 1.

The trillionth digit is given by the
(1000000000000 - (9 * 10987654321))%11 th digit of the
(1000000000000 - (9 * 10987654321))/11 11 digit number (which is 81919191919) or 1.

In general, the Nth digit of the seqence is given by
this digit: (N - (9 * S(d-1)))%d
of the (N - (9 * S(d-1)))/d 'th d digit number.
Where S(n) is the reverse of the original sequence (i.e. S(5) = 54321), and d is the biggest number such that (9 * S(d-1)) < N.

Hmmm. I just tried this for N = 1e18 and it didn't work, and I know why. Apparently by "in general" I meant "in general for N < ~1e10"

my reverse sequence is built from the fact that you can count the number of digits in the original sequence like this:
for N = 1e6 there are:
9 numbers with 1 digit
90 numbers with 2 digits
900 with 3, etc

total digits = 9*1 + 90*2 + 900*3 ...
= 9 * (1 + 10*2 + 100*3 ...)
= 9 * (...321)

but after 9 digit numbers this isn't true anymore.

sorry for the confusion.

One way to get a better algorithm is simplifying 1/9*(1-10^n+9*n*10^n),or find a simpler formula for biggest positive integer n,satisfies 1/9*(1-10^n+9*n*10^n) < = t (t- position number of unknown digit). May be it is possible to calculate short formula for some special values of t=10^s (as in problem 40 at [URL]http://projecteuler.net/index.php[/URL].).

[QUOTE=VolMike;112884]A part of problem 40 at [url]http://projecteuler.net/index.php[/url].

Thus, we have that all digits of all numbers in range (1..185184 (cause the 100000 is the first 6-digits number)) are included in our sequence (no more then 10^6 digits). Number of digits of 185185 which lie in sequence is equal 511111-85185*6=1. Thus, the millionth digit of sequence is the first digit of 185185- which equals 1.[/QUOTE]

:smile:

You have it right Volmike.

To make it clearer, dividing by 6 we get the quotient 85,185 and the remainder 1, so the desired digit occupies the first place in the 85,186th. 6- digit number. Since the first 6- digit number is 100,000 the 85,186th is 185,185 whose first digit is 1.

Thanks for the website. The next time I submit a problem I will check it and give something that is not included there!

Mally :coffee:

[quote=mfgoode;112934]:smile:

You have it right Volmike.

To make it clearer, dividing by 6 we get the quotient 85,185 and the remainder 1, so the desired digit occupies the first place in the 85,186th. 6- digit number. Since the first 6- digit number is 100,000 the 85,186th is 185,185 whose first digit is 1.

Thanks for the website. The next time I submit a problem I will check it and give something that is not included there!

Mally :coffee:[/quote]

Hm.. It seems that you (and others) post problems not only because you really need the answer, but to "shake up" the forum and make some kind of competition? If so, I start like this forum more and more :)

A Shake Up!
 

[QUOTE=VolMike;112937]Hm.. It seems that you (and others) post problems not only because you really need the answer, but to "shake up" the forum and make some kind of competition? If so, I start like this forum more and more :)[/QUOTE]

:smile:

You are dead right Vol. Speaking for myself yes I do. The aim is to test your knowledge and skill on important parts of Number theory and other topics
by putting it in the form of a puzzle.

You may be interested to know that in presenting a problem for the forum I came across an original theorem. I developed it as no one seemed to have solved it and have presented a paper for publication. I have also claimed that by it I have solved the Beal's conjecture. My paper awaits evaluation as it involves an enormous cash prize.

So much for puzzles. And I may add that many if not all of the great math'cians delighted in puzzles and challenged one another with tantalising ones. Well thats how Number theory developed!

Mally :coffee:

[quote=mfgoode;112974]:smile:

You are dead right Vol. Speaking for myself yes I do. The aim is to test your knowledge and skill on important parts of Number theory and other topics
by putting it in the form of a puzzle.

You may be interested to know that in presenting a problem for the forum I came across an original theorem. I developed it as no one seemed to have solved it and have presented a paper for publication. I have also claimed that by it I have solved the Beal's conjecture. My paper awaits evaluation as it involves an enormous cash prize.

So much for puzzles. And I may add that many if not all of the great math'cians delighted in puzzles and challenged one another with tantalising ones. Well thats how Number theory developed!

Mally :coffee:[/quote]

That's great! I hope we'll see your paper in Interrnet.

[QUOTE=VolMike;113022]That's great! I hope we'll see your paper in Interrnet.[/QUOTE]

:smile:

Thank you VolMike. I will be releasing it on the forum as soon as I have officially recorded it. It is very simple like the Pythagas theorem but true for all real and complex numbers. (I hope!)

Mally :coffee: