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MathGroup Archive 2005

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Re: Numbers and their reversals

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53714] Re: [mg53687] Numbers and their reversals
  • From: DrBob <drbob at bigfoot.com>
  • Date: Mon, 24 Jan 2005 03:37:43 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

There are actually LOTS of solutions.

Here's a verification of the ones you listed, including OTHER factors that work (other than 17 and 91). I'm not counting trivial cases where the product is its own reverse.

Clear[omariQ, omariD]
reverse = FromDigits@Reverse@IntegerDigits@# &;
commonDivisors = Rest@Divisors@GCD[#, reverse@#] &;
omariQ[n_Integer,
   k_Integer] := n != reverse@n && IntegerQ[n/k] && reverse[n/k] == reverse[n]/k
omariD[n_Integer] := Select[commonDivisors@n, omariQ[n, #] &]
omariQ[n_Integer] := Length@omariD@n > 0
omariD /@ {45628, 271726, 82654, 627172}
omariQ /@ {45628, 271726, 82654, 627172}

{{17,34,187,374},{91},{17,34,187,374},{91}}

{True,True,True,True}

And this counts the 5-digit numbers that can be factored that way:

Timing@Length@Select[Range[10000, 99999], omariQ]

{4.5 Second, 5887}

That's more than 6.5% of numbers in the range.

If we want only those for which 17 can be the common factor, here they are up to 5 digits:

Timing@Select[Range@99999, omariQ[#, 17] &]

{1.296 Second, {41140, 45628, 82280, 82654}}

91 yields quite a few more choices:

Timing@Select[Range@99999, omariQ[#, 91] &]

{1.25 Second, {10010, 11102,
     12012, 12103, 13013, 13104, 14014, 14105, 15015, 15106, 16016, 16107,
      17017, 17108, 18018, 18109, 20020, 20111, 21021, 21203, 22113, 22204,
      23023, 23114, 23205, 24024, 24115, 24206, 25025, 25116, 25207, 26026,
      26117, 26208, 27027, 27118, 27209, 30030, 30121, 30212, 31031, 31122,
      31304, 32032, 32214, 32305, 33124, 33215, 33306, 34034, 34125, 34216,
      34307, 35035, 35126, 35217, 35308, 36036, 36127, 36218, 36309, 40040,
      40131, 40222, 40313, 41041, 41132, 41223, 41405, 42042, 42133, 42315,
      42406, 43043, 43225, 43316, 43407, 44135, 44226, 44317, 44408, 45045,
      45136, 45227, 45318, 45409, 50050, 50141, 50232, 50323, 50414, 51051,
      51142, 51233, 51324, 51506, 52052, 52143, 52234, 52416, 52507, 53053,
      53144, 53326, 53417, 53508, 54054, 54236, 54327, 54418, 54509, 60060,
      60151, 60242, 60333, 60424, 60515, 61061, 61152, 61243, 61334, 61425,
      61607, 62062, 62153, 62244, 62335, 62517, 62608, 63063, 63154, 63245,
      63427, 63518, 63609, 70070, 70161, 70252, 70343, 70434, 70525, 70616,
      71071, 71162, 71253, 71344, 71435, 71526, 71708, 72072, 72163, 72254,
       72345, 72436, 72618, 72709, 80080, 80171, 80262, 80353, 80444, 80535,
       80626, 80717, 81081, 81172, 81263, 81354, 81445, 81536, 81627, 81809,
       90090, 90181, 90272, 90363, 90454, 90545, 90636, 90727, 90818}}

Bobby

On Sun, 23 Jan 2005 02:02:17 -0500 (EST), F. omari <towtoo2002 at yahoo.com> wrote:

>
> 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
>
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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