Pandigital ABCs (Posted on 2009-12-28)

	Submitted by Dej Mar
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Solution:	(Hide)
For AB = C : 48 (34105,    7,  8629) (13074,    8,  5692) (21547,    8, 99063) ( 5730,   12,  9684) ( 3672,   14,  9508) (  865,   39, 12407) (  568,   49, 12307) (  436,   52, 10978) (  706,   52, 18934) (  682,   54, 17930) (  701,   59, 24368) (  502,   63, 19847) (  802,   67, 35914) (  510,   69, 23874) (  360,   72, 15984) (  675,   83, 41920) (  304,   85, 21679) (  203,   86, 14795) (  270,   86, 15394) (  279,   86, 15403) (  205,   96, 18437) (  835,   96, 74021) (  340,   97, 28615) (   85, 1204,  9637) (   47, 1269,  5083) (   46, 1273,  5098) (   58, 1279,  6403) (   65, 1304,  7829) (   67, 1342,  8059) (   75, 1403,  9826) (   40, 1738,  6952) (   28, 1749,  3506) (   39, 1806,  5427) (   45, 1826,  7309) (   40, 1963,  7852) (   37, 2059,  6184) (   39, 2157,  6480) (   34, 2605,  7819) (   25, 3407,  6819) (   28, 3471,  6950) (   28, 3519,  7046) (   28, 3549,  7106) (   25, 3698,  7401) (   25, 4067,  8139) (   25, 4079,  8163) (   25, 4307,  8619) (   28, 4671,  9350) (   28, 4761,  9530) For AB = C : 0, as all integers is given to be greater than zero. If B were permitted to be 0 (as was my initial intent to allow), the value would be 40320 (8! permutations of the digits 2 through 9 for A, with B and C being 0 and 1, respectively).

	Subject	Author	Date
	computer solution to part 1	Charlie	2009-12-29 14:48:26
	re(2): Part 2: Trivial (?) solutions	Steve Herman	2009-12-29 10:24:23
	Part 2 solution	Justin	2009-12-28 19:53:52
	re: Part 2: Trivial (?) solutions	Justin	2009-12-28 19:33:34
	Part 2: Trivial (?) solutions	Steve Herman	2009-12-28 19:20:33