Coursera FB problem
Find 11 positive integers such that none of them add up to a multiple of 11, or prove that this cannot be done.
The above problem was posted on the FB page of the Coursera Introduction to mathematical thinking mooc.
To begin with I misunderstood the problem, however reading posts by other members of the group it became clear what the problem was and of how to proceed. This provided a very interesting experience of being in a virtual classroom.
I began the problem after my Japanese class on Tuesday night reading it on my iPhone in my car — please note that the car was stationary:) The final insight was provided by a fellow classmates posting several hours later as I got out of bed.
I am adding my attempt at understanding this problem below for the sake of interest. As discussed on the group I am looking at a smaller number to begin with ie showing the case for 3 positive integers:
 a, b, and c are +ve ints
 none of a, b, or c are divisible by 3
 hence a, b, and c leave remainder 1 or 2 when divided by 3
 we want to form the sums a+b, a+c, b+c, and a+b+c and to see what the remainder is when divided by 3
 it suffices to add the remainders of a, b, and c
 eg consider 4 and 8
 4 leaves remainder 1 when divided by 3
 8 leaves remainder 2 when divided by 3
 4 + 8 = 12 leaves remainder 0 when divided by 12
 however the remainders of 4 and 8 when we add them gives 1 + 2 = 3 but note that in terms of dividing by 3 a remainder of 3 is the same as a remainder of 0
 in mathematical notation we say 1+ 2 = 3 ≡ 0 (mod3)
 similarly consider 4 and 7, both leave remainder 1 when divided by 3
 4 + 7 = 11 which leaves a remainder of 2 when divided by 3
 note that the remainders of 4 and 7 are both 1 and that when we add these we get 2
a 
b 
c 
a+b 
a+c 
b+c 
a+b+c 

Remainder 
1 
1 
1 
2 
2 
2 
3 ≡ 0 (mod3) 
Remainder 
1 
1 
2 
2 
2 
3 
4 ≡ 1 (mod3) 
Remainder 
1 
2 
2 
3 
3 
4 
5 ≡ 2 (mod3) 
Remainder 
1 
2 
1 
3 
2 
3 
4 
Remainder 
2 
2 
2 
4 
4 
4 
6 ≡ 0 (mod3) 
Remainder 
2 
2 
1 
4 
3 
3 
5 
Remainder 
2 
1 
1 
3 
3 
2 
4 
Remainder 
2 
1 
2 
3 
4 
3 
5 
 notice that in each row there always occurs at least one cell in which there is a 3 or a 6 (≡ 0 (mod3) ie divisible by 3).
 with the necessary changes being made it would be possible to show that for any set containing n positive integers there must be at least one way of combining the numbers in the set to form a sum which is divisible by n.