In: Advanced Math
Let 
 be any 
 integers.
Case I: If some 
 divides one of 
, say 
, then the set that contains 
 has 
 as its subset, and the sum of all elements in 
 is just 
, a multiple of 
.
Case II: Suppose 
 for all 
: Since 
 sets contains 
 integers, by Pigeon-hole principle, at least one of these sets
contains at least 
 of these integers 
. Say 
 is such a set and 
. Since 
 for all 
, we have 
. Thus, the sum of the elements in the subset 
 is a multiple of 
.
Case III: Suppose 
 for all 
: Since 
 sets contains 
 integers, by Pigeon-hole principle, at least one of these sets
contains at least 
 of these integers 
. Say 
 is such a set and 
. Since 
 for all 
, we have 
. Thus, the sum of the elements in the subset 
 is a multiple of 
.
Case IV: Suppose 
 for at least one of 
, and 
 for at least one of 
:
Case IVa If one of the 
 sets contain an 
 and 
, then it has a two-elements subset 
 whose elements sum is 
.
Case IVb Since 
 sets contains 
 integers, by Pigeon-hole principle, at least one of these sets
contains at least 
 of these integers 
. Say 
 is such a set and 
. Either 
 for all 
, or 
 for all 
. Thus, we have 
. Thus, the sum of the elements in the subset 
 is a multiple of 
.
Since these are all the possible cases, we have proved the required statement.