Hi, I'm Vladimir Poldoskii,

and today we're going to discuss the properties of numbers to be even or odd.

Even numbers are integer numbers that are divisible by two.

And odd numbers are all other integer numbers.

For example, numbers 2, 4, 6,

8, and so on are even. They're divisible by two.

And numbers 1, 3, 5,

7, are odd, they're not divisible by two.

What about number zero?

Zero is also an even number, it is divisible by two.

We can divide zero by two and obtain zero.

It is also an integer,

so zero is even.

It is also useful to look at numbers on the number line.

Let's position all numbers in the increasing order, all integer numbers.

And let's color even numbers in green color and odd numbers in the red color.

Now we can see that odd and even numbers are interchangeable on this line.

Zero is even and one is odd and two is even and three is odd and so on,

and the same in reverse direction.

So even and odd numbers are interchangeable and this is good to keep in mind.

The properties of numbers to be even or odd

are important invariants that I use very often.

Let's see it on the example.

Consider a full on problem.

We have a chessboard and a piece which can move in one move,

in one move it can move to one cell that is adjacent by edge to the current one.

It can one cell the right,

to the left, up, or down.

Can it return to the original position after 17 moves or after 18 moves?

So there are two questions in this problem.

So let's proceed to a solution and let's start with a more simple case of 18 moves.

This case is more simple because it's possible to

return to the initial position in this case

and it is usually more simple to show

possibility to do something than impossibility to do something.

Okay, let's see how to do this.

It turns out that you can return to the initial position just after just two steps.

And you can do for example in this way,

you can move to the right first and then on the second step we can move

back and so they are back in the initial position after just two steps.

And now we can just repeat it nine times and

we'll be back in the initial position after 18 moves.

And note that here in the solution we used,

that 18 is an even number.

It is divisible by two,

so we can return to the initial position just two steps and then it

repeats a certain number of times and get to the initial position two after 18 moves.

So now let's proceed to the second part.

The previous solution doesn't work because 17 is odd,

and we can not just apply the same argument we used that 18 is even.

So now we have to do something else and it turns out that even more of it,

in this case this is impossible.

And let's see why.

It turns out that we can make a full on observation.

After even number of steps this piece is always on the white field.

And after odd number of steps it is always on the black field.

Indeed it, initially zero steps,

it is on the white field.

Then it moves after the first steps,

the number of steps becomes odd and the piece should move to the black cell,

and then after a second step it moves to the white cell.

But now the number of steps is two and it is even, and so on and so forth.

So we can observe it this also,

this property always is satisfied,

and this is our invariant.

And this means that there is no way to get back after odd number of steps.

Let's observe 17 as odd again.

Because after 17 number of steps you will be on

the black square and our initial square is white.