Bill It's impossible if they have to be distinct up to isomorphism, but otherwise this is a solution: 1: {e} 2: C_2 2: S_2 3: C_3 3: A_3 4: C_4 4: C_2 x C_2 5: C_5 6: C_6 6: S_3 6: D_6 7: C_7 8: Q_8 Total number of elements = 57. (where C_n = cyclic group of order n; D_2n = dihedral group of order 2n; S_n = symmetric group of order n!; A_n = alternating group of order n!/2; Q_8 = quaternions {±1, ±i, ±j, ±k} under multiplication; x denotes (exterior) direct product.) Sincerely, Adam P. Goucher http://cp4space.wordpress.com

----- Original Message ----- From: Bill Gosper Sent: 05/24/13 05:08 AM To: math-fun@mailman.xmission.com Subject: [math-fun] group theory

Yasmine wanted to divide 57 buttons into 13 groups. How many groups will Yasmine have? Will there be any buttons left over? (Saxon Math, Intermediate 4) --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun