Suppose there are 51 dalmatians and number of dots on each dalmatian is not null. Prove (or dis-prove) there is always a grouping such that at least one group has total number of dots as multiple of 11.
I can easily prove the statement for 101 dalmatians by modulus 11, but have no idea how to start for 51 dalmatians.
Given 12 dalmations, write their number of dots as $x_1,x_2,\cdots,x_{12}$.
Then two of $0,x_1,x_1+x_2,x_1+x_2+x_3,\cdots, x_1+x_2+\cdots + x_{11}$ are congruent modulo $11$. But that means $x_{n+1}+x_{n+2}+\cdots x_{m}$ must be divisible by $11$ for some $n<m$.
We need $12$ so that the "other" group always contains at least one dalmation - $x_{12}$. But that depends on what you mean by a "grouping."