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.