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        riddles >> medium >> really seeing double
        
(Message started by: JohanC on Dec 31^st, 2009, 7:25am)

Title: really seeing double
Post by JohanC on Dec 31^st, 2009, 7:25am

A generalization of the double seeing riddle (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1262263119) in the easy forum:

Let A be an N-digit number zyxw... (z not 0) and B be the 2N-digit number zyxw...zyxw... formed by writing the digits of A twice.
What would be the smallest such number for which B would be a square?

Title: Re: really seeing double
Post by BenVitale on Dec 31^st, 2009, 8:22am

So, A is the n-digit number zyxw... and B is the 2n-digit number zyxw...zyxw...

B = A * (10^n + 1)

we need to find an n in order to make 10^n+1 a square.

the trick is to find the smallest n !

Title: Re: really seeing double
Post by ThudanBlunder on Dec 31^st, 2009, 10:03am

[hide]10²¹ + 1 = 7*7*11*13*127*2689*459691*909091, the smallest number of this form with a repeated factor less than 10

So let number to be squared equal this number divided by 7 = 142857142857142857143

142857142857142857143² gives the repeating number 2040816326530612249920408163265306122499
where 20408163265306122499 = (10²¹ + 1)/49 [/hide]

Edit: NOT. Rechecking, I see that 142857142857142857143² = 20408163265306122499020408163265306122499

As 49 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/nless.gif 10, this method doesn't work. :( Now if only 3 were a factor....

Title: Re: really seeing double
Post by Aryabhatta on Dec 31^st, 2009, 10:12am

10^n +1 is never a perfect square.

[hide]
10^n+1 = 2 modulo 3, while any perfect square is either 0 or 1 modulo 3.
[/hide]

(Note: I was responding to BenVitale. The fact above does not imply that there is no solution to this problem)

Title: Re: really seeing double
Post by Aryabhatta on Dec 31^st, 2009, 4:14pm

This probably works: [hide] A=13223140496 [/hide]

Title: Re: really seeing double
Post by ThudanBlunder on Dec 31^st, 2009, 4:52pm

on 12/31/09 at 16:14:22, Aryabhatta wrote:

This probably works: [hide] A=13223140496 [/hide]

Duh, it was staring me in the face. ::) And multiplying by [hide]16/121 (rather than 4/121 or 9/121)[/hide] ensures it is the right size.

Title: Re: really seeing double
Post by JohanC on Jan 1^st, 2010, 8:51am

on 12/31/09 at 16:14:22, Aryabhatta wrote:

This probably works: [hide] A=13223140496 [/hide]

Well done !

Title: Re: really seeing double
Post by JohanC on Jan 1^st, 2010, 9:38am

on 12/31/09 at 10:03:05, ThudanBlunder wrote:

[hide]10²¹ + 1 = 7*7*11*13*127*2689*459691*909091, the smallest number of this form with a repeated factor less than 10.
....
[/hide]

Indeed, lots of interesting ideas.
Just to wrap up, although it looks like you already found out:
[hide]- we need to start with a number of the form 10^N + 1 with at least one repeated prime factor
- the repeated prime factor has to be divided away an even number of times
- to get into the desired range (between 10^N-1 and 10^N), we need multiplication with some small square
- the smallest such N is 11, with 100000000001 = 11²*23*4093*8779
- with that number there is only 11² that can be divided away, leaving 826446281
- 826446281 is too small, needing padding with zeros to get a square: 82644628100826446281
- multiplying with 4 or 9 is still too small
- only when multiplying with 16 we get into the correct range: 1322314049613223140496
- also multiplying with 25, 36, 49, 81 and 100 leads to numbers of the desired form
- when multiplying with 121 or larger, the number gets too big[/hide]