Re: Numbers and their reversals

*To*: mathgroup at smc.vnet.net*Subject*: [mg53706] Re: Numbers and their reversals*From*: "Scout" <user at domain.com>*Date*: Mon, 24 Jan 2005 03:37:27 -0500 (EST)*References*: <csvia0$asu$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Let n * k = z Rev(n) * k = Rev(z) -------------------------------------------------------------- ris={}; (* list of results in the form {number n, constant k} *) Do[ If[Mod[n,10]¹ 0, (* no closing zeroes *) nr=FromDigits[Reverse[IntegerDigits[n]]]; Do[ z=k*n; zr=FromDigits[Reverse[IntegerDigits[z]]]; If[zr¹ z, (* no palindromes *) If[nr*kSzr,AppendTo[ris,{n,k}]] ],{k,2,99} (* first and last k *) ] ],{n,11,101} (* first and last n *) ]; Print[ris]; ----------------------------------------------------- That's all. ~Scout~ "F. omari" > > i want to investigate the following two equations: > a * const = z > a_Reversed * const = z_Reversed > where a, z, and their reversed form and const are all positive integers > ie such that: > 2684 * 17 = 45628 > 4862 * 17 = 82654 > 2986 * 91 = 271726 > 6892 * 91 = 627172 > it happened that many multipliers of 91 have such a property. > while the multipliers of 17 have only 5 cases in the interval of 1 to 3000 > the following code will investigate the multipliers of 17, to investigate > another number just replace 17. and you may increase the interval of > investigation. i am sure that my code is an old fashion one, please any > other ideas about a more functional code. > a = Table[i, {i, 1, 3000}]; zR = ""; aR = 0; z = ""; > Do[aR = ToExpression[StringReverse[ToString[a[[i]]]]]; > z = ToString[a[[i]]*17]; > zR = StringReverse[ToString[aR*17]]; > If[zR == z, Print[a[[i]]]], {i, 1, 3000}] > > 242 > 484 > 2442 > 2662 > 2684 > regards > >