Author |
Topic: really seeing double (Read 610 times) |
|
JohanC
Senior Riddler
Posts: 460
|
 |
really seeing double
« on: Dec 31st, 2009, 7:25am » |
 Quote  Modify
|
A generalization of the double seeing riddle 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?
|
|
 IP Logged |
|
|
|
Benny
Uberpuzzler
Gender: 
Posts: 1024
|
|
Re: really seeing double
« Reply #1 on: Dec 31st, 2009, 8:22am » |
Quote Modify
|
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 !
|
|
IP Logged |
If we want to understand our world — or how to change it — we must first understand the rational choices that shape it.
|
|
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
|
Re: really seeing double
« Reply #2 on: Dec 31st, 2009, 10:03am » |
Quote Modify
|
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
Edit: NOT. Rechecking, I see that 1428571428571428571432 = 20408163265306122499020408163265306122499
As 49  10, this method doesn't work.  Now if only 3 were a factor....
|
| « Last Edit: Dec 31st, 2009, 2:00pm by ThudnBlunder » |
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: really seeing double
« Reply #3 on: Dec 31st, 2009, 10:12am » |
Quote Modify
|
10^n +1 is never a perfect square.
10^n+1 = 2 modulo 3, while any perfect square is either 0 or 1 modulo 3.
(Note: I was responding to BenVitale. The fact above does not imply that there is no solution to this problem)
|
| « Last Edit: Dec 31st, 2009, 10:15am by Aryabhatta » |
IP Logged |
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: really seeing double
« Reply #4 on: Dec 31st, 2009, 4:14pm » |
Quote Modify
|
This probably works: A=13223140496
|
| « Last Edit: Dec 31st, 2009, 4:14pm by Aryabhatta » |
IP Logged |
|
|
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
|
Re: really seeing double
« Reply #5 on: Dec 31st, 2009, 4:52pm » |
Quote Modify
|
on Dec 31st, 2009, 4:14pm, Aryabhatta wrote:| This probably works: A=13223140496 |
|
Duh, it was staring me in the face.  And multiplying by 16/121 (rather than 4/121 or 9/121) ensures it is the right size.
|
| « Last Edit: Jan 1st, 2010, 7:47am by ThudnBlunder » |
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
JohanC
Senior Riddler
Posts: 460
|
|
Re: really seeing double
« Reply #6 on: Jan 1st, 2010, 8:51am » |
Quote Modify
|
on Dec 31st, 2009, 4:14pm, Aryabhatta wrote:| This probably works: A=13223140496 |
|
Well done !
|
|
IP Logged |
|
|
|
JohanC
Senior Riddler
Posts: 460
|
|
Re: really seeing double
« Reply #7 on: Jan 1st, 2010, 9:38am » |
Quote Modify
|
on Dec 31st, 2009, 10:03am, ThudanBlunder wrote:1021 + 1 = 7*7*11*13*127*2689*459691*909091, the smallest number of this form with a repeated factor less than 10.
....
|
|
Indeed, lots of interesting ideas.
Just to wrap up, although it looks like you already found out:
- we need to start with a number of the form 10N + 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 10N-1 and 10N), 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
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print