A062567 First multiple of n whose reverse is also divisible by n, or 0 if no such multiple exists.

1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 11, 48, 494, 252, 510, 272, 272, 216, 171, 0, 168, 22, 161, 696, 525, 494, 999, 252, 232, 0, 434, 2112, 33, 272, 525, 216, 111, 494, 585, 0, 656, 252, 989, 44, 540, 414, 141, 2112, 343, 0, 969, 676, 212, 4698, 55, 616, 171, 232, 767

Offset: 1

Author

Erich Friedman, Jul 03 2001

Keywords

• base
• nonn

Comments

a(81) = 999999999. 10^27-1 is a solution for a(3^5), but it may not be the smallest one. However, it seems likely (and perhaps easy to prove) that a(3^i) is 3^(i-2) "9"s, for i > 1. - Jud McCranie, Aug 07 2001

a(3^5)=4899999987<10^27-1 so Jud McCranie's conjecture "for n>1, a(3^n)=10^3^(n-2)-1 " is incorrect. I found a(3^n) for n<21; A112726 gives this subsequence. From the terms of A112726 we see that for n>4, a(3^n) is much smaller than 10^3^(n-2)-1. It seems that only for n=2,3 & 4 we have a(3^n)=10^3^(n-2)-1. - Farideh Firoozbakht, Nov 13 2005

Examples

			48 and 84 are both divisible by 12.
		

Crossrefs

Cf. A112725, A112726.

Programs

Mathematica

Block[{k = 1}, While[ !IntegerQ[k/n] || !IntegerQ[ FromDigits[ Reverse[ IntegerDigits[k]]]/n] && k < 10^5, k++ ]; If[k != 10^5, k, 0]]; Table[ a[n], {n, 1, 60}] (* Robert G. Wilson v *)
a[n_]:=(For[m=1, !IntegerQ[FromDigits[Reverse[IntegerDigits[m*n]]]/n], m++ ]; m*n);Do[Print[a[n]], {n, 60}] (* Farideh Firoozbakht *)

Extensions

Offset corrected by Sean A. Irvine, Apr 03 2023