January 4th, 2007, 04:44 PM  #1 
Newbie Joined: Dec 2006 Posts: 29 Thanks: 0  Reversing Digits
As I was thinking about our problem involving 1's, 0's and squares, I came up with an interesting problem: Let N be a natural number and R(N) be the number obtained when the digits of N are reversed. Question: For which numbers does sqrt(R(N^2))=R(N)? I wrote a computer program and it seems that the only such numbers are those composed entirely of 0's, 1's, 2's and 3's (i.e. 103, 110, 111, 120, 221, etc.  Although some such as 23 do not have this property) Is there a simple reason behind this? Are these the only numbers with this property (did I not check enough numbers)? Thanks, Cat 
January 4th, 2007, 06:38 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
You're looking for numbers with R(n^2) = R(n)^2. Let me think...

January 4th, 2007, 06:50 PM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
OK, I wrote a program to generate these numbers. First of all, this sequence is on Sloane's list  A110810, not that it has anything there (unfortunately). I calculated all members of this sequence up to ten million (3,111,101 and 10,000,000 are the last two members) and it appeared to match your observation, though I have of yet no proof. I imagine it is not too hard. 
January 5th, 2007, 04:07 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
I think this is related to the fact that 3^2 < 10 and 4^2 > 10, with 10 as the base you're reversing in.

January 5th, 2007, 11:08 AM  #5  
Newbie Joined: Dec 2006 Posts: 29 Thanks: 0 
I recently found an article that might provide some information about the problem: (This is from http://www.mathpages.com/home/kmath185.htm towards the bottom) Quote:
23=3+2*B Roots: B=3/2 Hence, 23 = K(B+3/2) Clearly K=2, so 23=2(B+3/2) Now squaring, we have 23^2=529=4[(B+3/2)^2]=4[B^2+3B+9/4] =4B^2+12B+9 Now here's where the problem comes in. To use the trick in the article where we write the number rev(N)=ck + ... + c2 B^(k2) + c1 B^(k1) + c0 B^k, we need all of the c's to be <10, so we will use the fact that 10B=B^2 to rewrite 23^2 as: 23^2=4B^2+12B+9=5B^2+2B+9 (=529 as desired) Hence, we may reverse the numbers to obtain rev(23^2)=9B^2+2B+5=925, whose square root is 30.41, not 32. If we go the other route, rev(23)=32, and use the inverse root idea, we have 32^2 = K[(B+2/3)^2]. We find that K=9, so that 32^2=9[(B+2/3)^2] = 9B^2+12B+4 = 1024. So it seems that the problem arises when one of the coefficients of B is greater than 10. This will always happen when we have a digit greater than 3. Do you agree with the above ? I think this could help with a proof. Thanks, Cat  
January 6th, 2007, 09:46 AM  #6 
Newbie Joined: Dec 2006 Posts: 29 Thanks: 0 
This would also mean that the exact set of numbers are those that have no carries when squared 
January 7th, 2007, 07:32 PM  #7  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
 

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