SOLUTION: suppose I added up a long list say thousands of natural numbers with the property that and even number of them are odd and an odd number of them are even. Can you determine whether

Algebra ->  Formulas -> SOLUTION: suppose I added up a long list say thousands of natural numbers with the property that and even number of them are odd and an odd number of them are even. Can you determine whether      Log On


   

Question 543272: suppose I added up a long list say thousands of natural numbers with the property that and even number of them are odd and an odd number of them are even. Can you determine whether the sum will be even or odd? Prove your answer
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Yes it's bound to be even.

PROOF:

Every even natural number can be written as 2p, where p is a natural number.

Every odd natural number can be written as 2q-1, where q is a natural number.

Suppose M and N are natural numbers and the list contains

the odd number 2M-1 of even natural numbers 2p1,2p2,...,2p2M-1 

and

the even number 2N of odd natural numbers 2q1-1,2q2-1,...,2q2N-1.   

Then the sum of the numbers in the list = 

sum%28%282p%5Bi%5D-1%29%2Ci=1%2C2M%29 + sum%28%282q%5Bj%5D%29%2Cj=1%2C2N-1%29 =

sum%28%282p%5Bi%5D%29%2Ci=1%2C2M%29 - sum%28%281%29%2Ci=1%2C2M%29 + sum%28%282q%5Bj%5D%29%2Cj=1%2C2N-1%29 =

2sum%28%28p%5Bi%5D%29%2Ci=1%2C2M%29 - 2M + 2sum%28%28q%5Bj%5D%29%2Cj=1%2C2N-1%29

2%28sum%28%28p%5Bi%5D%29%2Ci=1%2C2M%29+-+M+%2B+sum%28%28q%5Bj%5D%29%2Cj=1%2C2N-1%29%29

which is an even number since it has a factor of 2.

[Note: We used 2p-1 for odd numbers rather than the usual 2p+1 because since
p is a natural number, the smallest odd natural number which 2p+1 can be is
3 when p=1, so to include 1, and some of the numbers in the list could be 1,
we must use 2p-1. The final result is positive since the first summation in
the parentheses above is greater than M.]

Edwin