1. odd + even = odd still, because even never changes it.

2. odd + odd = even because one of the odds changes it.

3. even + even = even because even doesn't change the other one, it skips by 2's.

Then I wondered what would happen if I included the answer, so I

started checking that out, and whammo!! I found one sentence that

satified all the conditions:

1. If you count the answer too, then the number of odds will always be even in a summation!!

Clever, huh? Just one sentence, and no inferences are needed.

Then you can derive backwards too from this to the original statements

by logically thinking it through.]]>

Just guessed. Anyway, what I did was prove why it will always work. Thought you were looking for an explanation.

]]>I thought he didn't see that there are always an even number if you count the answer,

so I reexplained that.]]>

What MIF is saying is if you have an even number of odds in your list the answer will always be even, post#3.

If you an odd number odds in your list than then the answer will be odd. So when you count up up all the odds and then aswer it will be odd + 1 = even. Post #1

]]>But the evenness even works for this list that comes out even.

For example.

13

15

18

+

____

46

Now we can see that their are still and even number of odd numbers: 13 and 15.

Notice the answer is included, but is not odd, so does not add one to the count.

So there will always be an even number of odd numbers if you

include the answer.]]>

list of integers, and get the correct

answer, then if you count all the integers

up that are odd including the answer if

it is odd, then the number of odd integers

will always be even!! The trick to

this is that the answer must be included.

Here's an example:

18

17

13

10

11

+

____

69

Now count up the odd integers:

17, 13, 11, and 69, that's four, an even number of odds!!

Cool, huh?

I thought this up while talking to my Mom today.]]>