SOLUTION: Prove that given any set of $17$ integers, not all odd, there exist nine of them whose sum is divisible by $2.$

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Question 1208373: Prove that given any set of $17$ integers, not all odd, there exist nine of them whose sum is divisible by $2.$

Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
The sum of any number of evens is even.
The sum of any even number of odds is even. 

Case 1:
If there are 9 or more evens, then the sum of 9 evens will be even.

Case 2:
If there are only 8 or fewer evens (but at least 1 even), then there are 9 or 
more odds.  
Take 8 of those odds and 1 even, and the sum of those 9 will be even. 

PROVED.

Edwin
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Prove that given any set of 17 integers, not all odd, there exist nine of them whose sum is divisible by 2.
~~~~~~~~~~~~~~~~~~~~~~~ 

Let m be the number of odd  integers in our sets of 17 integer numbers, and

let n be the number of even integers in our sets of 17 integer numbers.


We have m + n = 17.


        It implies that  EITHER  m >= 9  OR  n >= 9.


In other words, one of the two integer numbers, m or n, must be at least 9.


Indeed, otherwise the sum  m+n  would not be more than  8 + 8 = 16;  but m+n = 17.



If n >= 9, then we can take 9 even integer numbers from our set.  

           Their sum will be divisible by 2, so in this case these 9 integer numbers are the seeking set.



If m >= 9, then there are two sub-cases:

           - (a)  all integer numbers in our set are odd.  
                  It contradict to the imposed condition,  so this case can not happen.


           - (b)  there is at least one even number in our set of 17 integer numbers.

                  In this case, we form the set of 9 numbers, taking 8 odd numbers and this even number.

                  The sum of these 9 integer numbers is even number.


So, we proved that in any case, it is possible to find a subset of 9 integers with even sum.

The proof is complete and the problem is solved.



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