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Topic: Update: string of digits repeated (Read 3226 times) |
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Christine
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Update: string of digits repeated
« on: Nov 9th, 2012, 10:24am » |
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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.
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| « Last Edit: Nov 9th, 2012, 12:03pm by Christine » |
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towr
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Re: Update: string of digits repeated
« Reply #1 on: Nov 9th, 2012, 12:48pm » |
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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
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| « Last Edit: Nov 9th, 2012, 12:55pm by towr » |
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towr
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Re: Update: string of digits repeated
« Reply #2 on: Nov 9th, 2012, 1:44pm » |
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Case 1: 1322314049613223140496
Actually easier than I though, once you think about it.
Here's a longer one:
206611570247933884297520661157025206611570247933884297520661157025
Longer still!
297520661157024793388429752066115702479338842975206611629752066115702479 33884297520661157024793388429752066116
(10(22k+11)+1)2/121 * K2 with K=4..10
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| « Last Edit: Nov 9th, 2012, 1:51pm by towr » |
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Christine
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Re: Update: string of digits repeated
« Reply #3 on: Nov 9th, 2012, 1:49pm » |
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on Nov 9th, 2012, 12:48pm, towr wrote: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 |
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Case #1 is not possible
abcdabcd = 10001 * abcd
10001 = 73 * 137
the prime decomposition of abcdabcd must contain factors to an even power
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Christine
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Re: Update: string of digits repeated
« Reply #4 on: Nov 9th, 2012, 1:51pm » |
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on Nov 9th, 2012, 1:44pm, towr wrote:Case 1: 1322314049613223140496
Actually easier than I though, once you think about it.
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how did you get it?
Quote:
Here's a longer one:
132231404958677685950413223140496132231404958677685950413223140496
Longer still!
132231404958677685950413223140495867768595041322314049613223140495867768 59504132231404958677685950413223140496
(10(22k+11)+1)2/121 * 16
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towr
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Re: Update: string of digits repeated
« Reply #5 on: Nov 9th, 2012, 1:54pm » |
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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).
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| « Last Edit: Nov 9th, 2012, 2:02pm by towr » |
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Christine
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Re: Update: string of digits repeated
« Reply #6 on: Nov 9th, 2012, 2:27pm » |
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Thank you towr
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towr
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Re: Update: string of digits repeated
« Reply #7 on: Nov 9th, 2012, 2:32pm » |
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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 10n-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..
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| « Last Edit: Nov 9th, 2012, 2:38pm by towr » |
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