22) Consider the set $\{1,2, \ldots, 2003\}$. How many of its subsets $B$ are there such that the sum of the elements of $B$ is equal to 2007000?
(A) None exist
(B) 3
(C) 4
(D) 1002
(E) 2003.
22) The answer is (C). The sum of the integers from 1 to $n$ is equal to $\frac{n(n+1)}{2}$, and thus the sum of all integers from 1 to 2003 is equal to 2007006. It follows that a subset of $\{1,2, \ldots, 2003\}$ is such that the sum of its elements is equal to 2007000 if, and only if, the sum of the elements of its complement is equal to 6. It is therefore equivalent to count these latter. They are evidently only $\{1,2,3\},\{1,5\},\{2,4\}\{6\}$.