Which of the following are complete residue systems modulo $11$? Which of the following are complete residue systems modulo 11?
$(a)\quad  0,1,2,4,8,16,32,64,128,256,512$
$(b)\quad  1,3,5,7,9,11,13,15,17,19,21$
$(c)\quad  2,4,6,8,10,12,14,16,17,20,22$
$(d)\quad  -5,-4,-3,-2,-1,0,1,2,3,4,5$
I have the answer which is that they are all complete residue systems modulo 11. However, I do not fully understand the definition of residue system or how to determine what sets make up a complete residue system?
 A: To be a complete residue system modulo $n$, you need a set of $n$ integers, no two of which are congruent modulo $n$.
A: The set of integers {0, 1, 2, ..., n - 1} is called the least residue system modulo n. Any set of n integers, $\bf no\ two\ of\ which\ are\ congruent\ modulo\ n$, is called a $\bf complete\ residue\ system\ modulo\ n$.
http://en.wikipedia.org/wiki/Complete_residue_system_modulo_m#Residue_systems
{0,1,2,3,4,5,6,7,8,9,10} is a complete residue system modulo 11. 
Since $1 \equiv 12 \pmod{11}$,$3 \equiv 14 \pmod{11}$, ... , $10 \equiv 21 \pmod{11}$.
After dividing 11 for each of them, the residues are:
(a)0,1,2,4,8,5,10,9,7,3,6;
(b)1,3,5,7,9,0,2,4,6,8,10;
(c)2,4,6,8,10,1,3,5,6,9,0; (c isn't)
(d)6,7,8,9,10,0,1,2,3,4,5.
A: Picture a 12-hour clock where we only count $1,2,4,5,6,7,8,9,10,11,12$ and hours. Would this clock be any good?
No because there is no number that measures times when we are $3$ hours past midday/midnight.
How about if we use the numbers $1,14,-8,125,6,31,-16,9,22,0$? No, again we still don't count the times that are $3$ hours past.
This is the point of a residue system, we want a set of numbers that DO give every possibility mod $N$. We would also like no repetitions too, hence a complete residue system.
A: Any set of n integers, no two of which are congruent modulo n, is called a complete residue system modulo n. In all of these four options, you have n=11 elements and for some x,y belongs to this set are incongruent. i.e x is not congruent to y mod n. It means (x-y) is not divisible by n. 
note :- x,y are the elements of the set you provided.
A: CRS for a divisor means all remainders possible with the divisor when any integer is divided by the divisor. The CRS includes 0 which means divisor has completely divided the integer leaving 0 residue. Thus if divisor is 8: possible remainders with respect to any dividend are 0, 1, 2, 3, 4, 5, 6, 7 (total 8). The given set is considered CRS if it includes all 8 divisors for this test divisor.  The set would contain any multiple of divisor with some residue of its CRS. This can be simply checked by dividing each element by 8 and listing what is remainder (residue). If set includes integers less than the divisor (instant case 8) they are taken as such; only integers larger than the divisor are divided for finding remainder. If all residues when arranged in order match expected list; the system is declared having CRS. Nutshell: every divisor classifies all integers as its own system having number of classes equal to its CRS because the entire integer system is multiple of the divisor leaving residue as per the expected class from 0 through n-1 (if n is the divisor).
A: The correct answer is (b). 1, 3, 5, 7, and 9 are in their own remainder classes mod 11 while 11 is in cl(0), 13 in cl(2), 15 in cl(4), 17 in cl(6) 19 in cl(8), and 21 in cl(10). We have members of every remainder class  listed hence a complete residue system. The key is to know that all we need for a complete residue  system are members of every remainder class. Thomas defines a complete residue class as a set $S=\{r,s,\dotsc\}$ such that for every integer $k$ there exists a member of $S$, call it $m$, such that $k$ is congruent to $m$ mod $n$ or in the problem at hand mod 11. The complete residue system partitions the set of integers. It's important to realize that complete residue systems are not unique. All that's needed is that the set (the complete residue class) include any member of every remainder class.
