UPDATE (5 May): Here is another basic theorem in group theory
you might find useful. If G is a finite group and H is a subgroup
of it (a subset of the elements, with the same group operation, that also
forms a group), then |H| divides |G|. This is known as Lagrange's
theorem, and is most often used when H is the group
{1, x, x^{2},...} for some x in G.
This month's riddle comes from Serge Gautier. (Thanks, Serge!) It extends one of the very earliest Using your Head is Permitted riddles (December 2007), which is a personal favourite of mine, and one that has already been extended once (in June 2010).
A **Associativity**: For every*x*,*y*,*z*in*S*,*x**(*y***z*)=(*x***y*)**z*.**Identity**: There exists an element,*i*in*S*, such that for every*x*in*S*,*x***i*=*i***x*=*x*. This is known as the group's identity element. (It is necessarily unique.)**Inverse element**: For every*x*in*S*there exists a*y*in*S*such that*x***y*=*y***x*=*i*, the identity element.
A group is called Traditionally, the group operation of abelian groups is denoted "+". We now repeat the December 2007 riddle:
Let
## Part 1:For which values ofn is it possible to find such A,
B and C satisfying that for all x in S,
A(x)+B(x)=C(x)?
## Part 2:For which values ofn is it possible to find such A,
B and C satisfying that for all x in S,
A(x)*B(x)=C(x)?
The only difference from the original riddle is that now we are asking this
question not only over standard multiplication and standard addition, but over
In other words, you are to answer whether for each possible choice of As always, prove your answer.
A separate list of solvers will be kept for each of the two parts.
Answer |
## List of solvers:## Part 1:Radu-Alexandru Todor (1 May 08:58)JJ Rabeyrin (1 May 20:14) Harald Bögeholz (3 May 12:24) Uoti Urpala (3 May 12:38) Lian Wang (3 May 19:06) Hanlin Ren (3 May 21:33) Jens Voss (4 May 16:07) Dan Dima (5 May 03:52) Lin Jin (9 May 14:06) Oscar Volpatti (30 May 00:02) ## Part 2:Uoti Urpala (3 May 12:38)Lian Wang (3 May 19:06) Hanlin Ren (3 May 21:33) Jens Voss (4 May 16:07) JJ Rabeyrin (8 May 07:34) Lin Jin (9 May 14:06) Harald Bögeholz (23 May 10:50) |

Elegant and original solutions can be submitted to the puzzlemaster at __riddlesbrand.scso.com__.
Names of solvers will be posted on this page. Notify if you don't want
your name to be mentioned.

The solution will be published at the end of the month.

Enjoy!