If you assume abc is the smallest number of each triplet, the solutions are a=2,b=4,c=3 and a=5,b=6,c=7. The former solution is the one that yields a sixth power. Pretty clever. I solved this one with a combination of a hand calculator, logical reasoning, and brute effort. Darrel C Jones abc=243 or 567 I like ur page. Nils Mellech Well, not really. It seems to me that the number of solutions must be a multiple of 3, since if abc is a solution then so must be bca and cab. But I understand what you meant. Anyhow, abc = 243 works (243×432×324 = 34,012,224 = 5832^2 = 18^6) as does abc = 567 (567×675×756 = 289,340,100 = 17010^2). Al Zimmermann The only two solutions to Robert Reid's problem ("find a three-digit number abc such that abc*bca*cab is a square number") up to rotation are 234 * 342 * 423 = 34012224 = 5832^2 ( = 324^3 = 18^6) 576 * 765 * 657 = 289340100 = 17010^2 Considering higher powers (not just squares) yields, in addition to the solutions above and the obvious cases a=b=c (abc * bca * cab = aaa^3), one additional solution: 468 * 684 * 846 = 272097792 = 648^3 Those are the only solutions to abc * bca * cab = N^k for k > 1. I also looked at numbers with more digits, such as abcd*bcda*cdab*dabc and abcde*bcdea*cdeab*deabc*eabcd, but these did not yield solutions other than the obvious ones: Michael Brundage { abc , bca , cab : abc X bca X cba is a perfect square } = { 243 , 432 , 324 } ( Square 34012224 = 5832 ^2 ) = { 567 , 675 , 756 } ( Square 289340100 = 17010 ^2 ) Sudipta Das Given the language of your original challenge: "Find a three digit number abc such that abc×bca×cab is a square number." there are, pedantically speaking, six solutions: abc = 432, abc = 324, abc = 243, abc = 567, abc = 675, abc = 756 However there are only two unique squares. The 432, et.al. group are the ones that generate the sixth power, who's root is 18. Thanks for the fun challenge! Mike Kurtinitis Robert Reid's three digit number is: 1: 243x432x324 = 34012224 = 5823^2 2: 567x675x756 = 289340100 = 17010^2 --Koshi Arai Solved using Perl 243 x 324 x 432 = 5832^2 567 x 675 x 756 = 17010^2 #!/usr/bin/perl use warnings; use strict; my (\$a, \$b, \$c); my (\$p, \$q, \$r); my (\$t); for (my \$x=100; \$x<1000; \$x++) { \$a=substr(\$x,0,1); \$b=substr(\$x,1,1); \$c=substr(\$x,2,1); \$p=int(\$a.\$b.\$c); \$q=int(\$b.\$c.\$a); \$r=int(\$c.\$a.\$b); \$t=sqrt(\$p*\$q*\$r); if (length(\$t) eq length(int(\$t))) { print "\$a \$b \$c", "\n"; } } exit; Michael J D Dufour BSc. (Hon.) MSc. Nice to have an easy one for once! 243 x 432 x 324 = 5832^2 = 18^6 567 x 675 x 756 = 17010^2 John Gowland The answer for abc are 243 and 567. abc=243 gives abc*bca*cab as 6th power i.e. 18^6 Shyam Sunder Gupta 1. 243 2. 567 Jared Marks James Lewis Melby Matt Coury I solve a lot of puzzles from this site and I enjoy almost every puzzle. But I don't send the answers. A friend of mine suggested that I should send them, so I start with this easier one. This site is Really a great work and thanks a lot for each and every puzzle that I enjoyed. 1. abc = 243 (243 x 432 x 324 = 5832 squared = 18^6) 2. abc = 567 (567 x 675 x 756 = 17010 squared) Anupam 234 x 243 x 432 = 34012224 = 5832 x 5832 567 x 576 x 756 = 289340100 = 17010 x 17010 by Javier Arbonés (Barcelona-Spain)