Positive integers i, j, k, m, and n are randomly chosen (repetition is allowed) so that
2 <= i,j,k,m,n <= 2009 . What is the probability that ijk + mn is even?
The Attempt at a Solution
Each of the positive integers is one out of 2008 numbers. 1004 of these numbers are even and 1004 of these numbers are odd. So, 50% chance of being an even number and 50% chance of being an odd number for each of the integers i,j,k,m,n.
I looked at the ijk term first and wrote out all possible combinations.
odd x odd x odd = odd
odd x odd x even = even
odd x even x odd = even
odd x even x even = even
even x odd x odd = even
even x odd x even = even
even x even x odd = even
even x even x even = even
1/8 possibilities are odd (12.5%) and 7/8 are even (87.5%)
I looked at the mn term next and did the same.
odd x odd = odd
odd x even = even
even x even = even
even x odd = even
1/4 possibilities are odd (25%) and 3/4 are even (75%)
Next, I looked at the term ijk and the term mn added together.
odd + odd = even
odd + even = odd
even + odd = odd
even + even = even
2/4 possibilities are odd (50%) and 2/4 are even (50%)
I'm really not sure where to go from here. Am I on the right track at least? I'm just not sure how to combine the above facts into a statement about the probability of [ijk + mn] being an even number. Any help is much appreciated