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 January 4th, 2007, 05:44 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, 07:38 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, 07:50 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, 05:07 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, 12:08 PM   #5
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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:
 Originally Posted by Math Pages There is, however, validity in the argument that some problems are "natural" whereas others are mere contrivances. For example, it might be argued that the "digital reversal" of a number is not a natural or meaningful concept, but consider a number N with the base-B representation c0 + c1 B + c2 B^2 + ... + ck B^k What are the "roots" of this number? In other words, given the coefficients c0,c1,...ck, for what values of B does the above expression equal zero? Now consider the digit reversal of N: ck + ... + c2 B^(k-2) + c1 B^(k-1) + c0 B^k What are the roots of this number? If we normalize these two polynomials the roots of N are simply the inverses of the roots of rev(N). Thus, rev(N) can be expressed naturally in terms of the roots of N (and vice versa).
I decided to play around with this using the number 23 which we know is not one of our numbers (from this point on, B=10):

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^(k-2) + c1 B^(k-1) + 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, 10:46 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, 08:32 PM   #7
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Quote:
 Originally Posted by Cat 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.
I agree that carries seem to be the problem -- as I mentioned, I think, in one of my posts above. I'm not sure how to turn this into a proof, though. Certainly it doesn't seem hard to show that if there are no carries the number is of the indicated form, but showing that that form is the only one possible seems harder. Tags digits, reversing Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Vidofner Applied Math 4 January 12th, 2014 09:47 AM oneup Elementary Math 4 September 2nd, 2013 05:08 AM CarpeDiem Elementary Math 7 July 13th, 2012 08:35 PM Liqwde Computer Science 2 August 31st, 2010 09:59 PM Gotovina7 Calculus 1 February 29th, 2008 10:30 AM

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