A set of 7 integers has a range of 2, an average of 3, and a mode of 3. 
A math question asks a question by providing information: "A set of $7$ integers has a range of $2$, an
  average of $3$, and a mode of $3$"

My questions is how can a set of $7$ integers have a range $2$ ? As I know elements can't be repeated in a set. Then, if if I make a list of numbers as {2,2,3,3,3,4,4}, which is one possibility to fulfill above conditions, is it a set of 7 integers?
 A: I think that the author of the question meant "multiset" rather than "set". This kind of question has appeared on the GMAT in the past (as a quick Google search will reveal, see one example: http://gmatclub.com/forum/a-set-of-numbers-contains-7-integers-and-has-an-average-132599.html) and the answers usually consist of multisets.
A: You are right, a set cannot have repeated elements. Probably the writer of the question is using a looser definition meaning 'collection' or 'multiset'. They shouldn't, I am just guessing what might have happened. 
Also note that the answer you are looking is {2,2,3,3,3,4,4} not {1,1,3,3,3,4,4}. Maybe you just made a typo(?)
Can you prove that this is the only answer? Or can you find all different forms that this collection has? If you have difficulties with the proof comment below and I'll edit my answer to include it.
EDIT: Here's the proof. 


*

*3 is the mode, so it should be in the collection (we don't know how many times yet).

*There should be numbers other than 3, since the range of the collection is 2 (and not 0).

*For any number above 3, we should have a corresponding number below 3, because the average is 3. So if number 4 is included then number 2 must be included too.

*2 and 4 are the only numbers that can be included in the collection. Any number above 4 means that a number below 2 must be included with makes the range > 2.

*So the set includes 2 (n times), 3 (m times), and 4 (n times). 2 and 4 must occur equal times if we want the average to be 3.

*So we are looking for positive integers $n,m$ such that $n + m + n = 7$ and  $m > n$ (since 3 is the mode). For $n=1$ we get $m=5$. For $n=2$ we get $m=3$. For any $n>2$ we get a $m<n$ so this cannot be a solution. 


Thus the only solutions are {2,3,3,3,3,3,4} and {2,2,3,3,3,4,4}
