Abstract: | Let A N. The cardinality (the sum of the elements) of A willbe denoted by |A| ((A)). Let m N and p be a prime. Let A {1, 2,...,p}. We prove thefollowing results. If |A| (p+m2)/m]+m, then for every integer x such that0 x p 1, there is B A such that |B| = m and (B) x mod p. Moreover, the bound is attained. If |A| (p+m2)/m]+m!, then there is B A such that |B| 0 mod m and (B) = (m 1)!p. If |A| (p + 1)/3]+29, then for every even integer x such that4p s x p(p + 170)/48, there is S A such that x = (S). In particular,for every even integer a 2 such that p 192a 170, thereare an integer j 0 and S A such that (S) = aj+1. |