pigeon hole different solution

Question

Using pigeon hole ,to prove that "whatever" triad of natural numbers, have at least one pair of numbers, those sum must be even.

What I did, on my way I don't know if it is wrong what I did.Maybe yes, I would like to tell me.

I take 3 numbers  example 2 , 4 , 1 . After all those numbers, I add them 2+4+1 =7 .
I take lets say 10 and I did Pigeon hole [10/7]=[1.4]=2 .

This is completely wrong right?

Answer 1

Hint: Numbers are either even or odd. If two numbers are the same they add to an even. To add to an odd the numbers have to be different.

So given a choice of even/odd, prove that at least $2$ of the $3$ numbers are the same. (They can not all be different from each other.)

Answer:

There's only $2$ types of numbers and we have $3$ numbers. So have more numbers than types of numbers so at least two numbers must be the same type. So those two must add to an even number.

Answer 2

basic idea is this ...

even odd

are containers,

num1,num2,num3

are objects

assume 1,2,4 then

even has 2,4

odd has 1

we can't change 2, to an odd number $x$ because then

even has 4

odd 1,x

now the sum of two odd numbers is even (2k+1)+(2j+1)=2(k+j+1)

and the sum of two even numbers is even (2k)+(2j)=2(k+j)

so since we no matter what we have a pair in at least one category, we always have a sum that is even.