anyone intelligent 2

working backwards....

ending cases( A →B )

A B
1 n-1 ( For A to have 1 either it must start or it must ve got it frm another person which is impossible as it wud ve to be 0 then implying it already left table) so A starts is only possible ie n=2
A B
2 n-2 ( here someone must give A 1 penny or B itself gives)

ie
1 ) C gives A
C A B
1 1 n-2 here again C shud start in first place implying n=3

2) B gives A

A B
1 n-1
back 1 step A C B
1 2 n-3
back 1 step A D C B
1 1 1 n-3
which implies D has to start implying n=4

hence we can see n=1,2,3,4 alone satisfy

and THERE IS NO SUCH INFINITE SET as mentioned in the question !!!!!!

no celestine. i am getting different. wait i will post my method

hey how did u get that? i did not get in terms of `n` !!!!!!!!!

should i post?????

ya surely!

here is the soln....any doubts plz ask.....

[1][1][1]

yes Eureka terrific sol !!!!!

i had messed a bit on backward reasoning and found the flaw in it :)