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        riddles >> easy >> Update: string of digits repeated
        
(Message started by: Christine on Nov 9^th, 2012, 10:24am)

Title: Update: string of digits repeated
Post by Christine on Nov 9^th, 2012, 10:24am

Can a number with a string of digits repeated be a square number?

A number of the form abcd...abcd...

I meant:

Case 1: Is it possible to find a 2n long square number with the following property: The number of the first n digits, and the number of the last n digits, in this order, are the same.

Case 2: Is it possible to find a 2n+k long square number. A number of the form abcdexyabcde, for example

I posted this question online, got no response.

Title: Re: Update: string of digits repeated
Post by towr on Nov 9^th, 2012, 12:48pm

For case 2, there's 121, although n=1,k=1 is perhaps a bit trivial.

Searching for increasing repeating lengths, I've so far found
1521
163216
12173121
1364711364
211568721156
13843653138436
1519521401519521

Title: Re: Update: string of digits repeated
Post by towr on Nov 9^th, 2012, 1:44pm

Case 1: 1322314049613223140496
Actually easier than I though, once you think about it.

Here's a longer one:
206611570247933884297520661157025206611570247933884297520661157025

Longer still!
29752066115702479338842975206611570247933884297520661162975206611570247933884297520661157024793388429752066116

[hide](10^(22k+11)+1)²/121 * K² with K=4..10[/hide]

Title: Re: Update: string of digits repeated
Post by Christine on Nov 9^th, 2012, 1:49pm

on 11/09/12 at 12:48:09, towr wrote:

Case #1 is not possible
abcdabcd = 10001 * abcd
10001 = 73 * 137
the prime decomposition of abcdabcd must contain factors to an even power

Title: Re: Update: string of digits repeated
Post by Christine on Nov 9^th, 2012, 1:51pm

on 11/09/12 at 13:44:56, towr wrote:

Case 1: 1322314049613223140496
Actually easier than I though, once you think about it.

how did you get it?

Quote:

Here's a longer one:
132231404958677685950413223140496132231404958677685950413223140496

Longer still!
13223140495867768595041322314049586776859504132231404961322314049586776859504132231404958677685950413223140496

[hide](10^(22k+11)+1)²/121 * 16[/hide]

Title: Re: Update: string of digits repeated
Post by towr on Nov 9^th, 2012, 1:54pm

I happen to know how to test for divisibility by 11: alternately add/subtract the digits if the sum is divisible by 11, then so is the number.
So, then I got to work on finding a square of the form (10²ⁿ⁺¹+1)*K, where the digits of the first and second half will cancel each other out.
10²ⁿ⁺¹+1 is divisible by 11, but if it's divisible by 121 that is even better, because you can square it and divide by 121 and you're almost there. But then it needs to be of the form (10²²ⁿ⁺¹¹+1).

Title: Re: Update: string of digits repeated
Post by Christine on Nov 9^th, 2012, 2:27pm

Thank you towr

Title: Re: Update: string of digits repeated
Post by towr on Nov 9^th, 2012, 2:32pm

I think 1322314049613223140496 is also the smallest such number, because we need a factor 10ⁿ+1, and as you suggested, this 10ⁿ+1 factor needs to be divisible by at least one square, otherwise we cannot complement each non-square among its factors with one from the number K (where 10^n-1 <= K <10ⁿ, i.e. such that (10ⁿ+1)*K = abcd...abcd... )
Pretty neat puzzle.

I suppose a followup question is if we can get any multiple repeats of abcd... Like, for example thrice: abcd..abcd..abcd..