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