Question 939395
  
Given:
a set A of 9 odd integers (whole numbers), A={1,3,5,7,9,11,13,15,17}
  
Objective: obtain a sum of 50 using exactly 5 elements of set A.

Analysis:
Assume it is possible to find 5 numbers n1...n5 such that their sum is 50.
Then
{{{n1+n2+n3+n4+n5=50}}}
Since all these numbers are odd, each can be represented by n1=2k1+1, where k1 is an arbitrary integer (whole number), and the same for n2...n5.
The sum is then
{{{2k1+1 + 2k2+1 + 2k3+1 + 2k4+1 + 2k5+1 = 50}}}
rearranging,
{{{2(k1+k2+k3+k4+k5) + 5 = 50}}}
however, it is evident that 2(k1+k2+k3+k4+k5) is an even number,
therefore 
{{{even number + 5 = 50}}} cannot be a true statement.
Thus we conclude that the initial assumption
{{{n1+n2+n3+n4+n5=50}}} is also a false statement.
  
Conclusion: There is no solution to the given problem, proved by contradiction.
  
Answer:
There is no solution to this problem because the sum of an odd number of odd numbers is always odd.