Count Your Marbles

At first the marbles are divided as :- Y and X

After first move marbles are divided as

Y + $\frac {X}{2}$ and X - $\frac {X}{2}$

After second move

$\frac {Y + \frac {X}{2} }{2}$ and $\frac {X}{2}$ + $\frac {Y + \frac {X}{2} }{2}$

That is

$\frac {2Y + X}{4}$ and $\frac {3X + 2Y}{4}$

Now it is given that either :--

$\frac {2Y + X}{4}$ = Y OR $\frac {2Y + X}{4}$ = X

So either :--

2Y = X OR 3X = 2Y

we need to equate the above result with X + Y = 2020

When 2Y = X then we get values of X And Y in fraction which is not possible

Thus 3X = 2Y, and on equating with X + Y = 2020 we get X = 808 and Y = 1212

According to the question we need to find marbles left with me which is $\frac {3X + 2Y}{4}$ = $\boxed {1212}$

We have $X+Y=2020$ . After the exchanges, the amount of marbles "I" have is $\frac Y2 + \frac X4 = X \text{ or } Y$ . $\frac Y2 + \frac X4 = Y \implies Y = \frac{2020}3 \notin \mathbb{N}$ , so it must be $\frac Y2 + \frac X4 = X \implies Y = 1212$ which is the amount of marbles "you" have when "I" have $X$ marbles.

X + Y = 2020 .....EQN(1)

FOR FIRST EXCHANGE,. X- X/2 I HAVE AND Y+X/2 YOU HAVE .

FOR SECOND EXCHANGE, X-X/2 +1/2(Y+X/2) I HAVE AND YOU HAVE Y + X/2 - 1/2(Y+X/2) .

SO NOW, YOU HAVE Y/2 + X/4 MARBLES. THIS MAY BE EQUAL TO X OR Y AS PER THE QUESTION.

WHEN Y/2 + X/4 = Y , WE GET X= 2Y . PUTTING THIS IN EQN(1) 3Y = 2020 BUT THE VALUE OF Y CAN'T BE FRACTION SO THIS ASSUMPTION IS WRONG.

WHEN Y/2 + X/4 = X , WE GET Y= 3X/2 . PUTTING THIS IN EQN(1) 5X/2 = 2020 OR X = 808 and Y = 1212.

SO YOU HAVE 1212 MARBLES.

This could only happen with a 60%-40% split. We then go to an 80%-20% split after half (20%) of the 40% gets added to the 60%. Lastly, we take half (40%) of the 80% and give it back to what became 20%. That leaves us with a 40%-60% split.

60% of 2020 is 1212. 40% of 2020 is 808. Since I (X) “gave” marbles first, I had to have started with the 40% first otherwise we would finish with a 35%-65% split if I had 60% of the marbles to start with. I started with 40%, 808, and ended with 60%, 1212.

As we know 2020 - Y = X, we could write the state at the end of the exchange as either

Option 1: ([2020 - Y]/2 + Y)/2 = Y

Or

Option 2: ([2020 - Y]/2 + Y)/2 = X = 2020 - Y

In the first scenario, as we solve for Y we conclude that

Y = 2020/3

This results in a non integer number of marbles (which is impossible). This does not happen in the second scenario, in which solving for Y gets us to a valid amount of marbles in our hands.

([2020 - Y]/2 + Y)/2 = 2020 - Y

[2020 - Y]/2 = 2(2020 - Y) - Y

2020 - Y = 2(4040 - 3Y)

5Y = 8080 - 2020

Y = 6060/5

Y = 1212

1) X + Y = 2020

2) X/2 + (Y + X/2) = 2020

3) [(X/2)+(Y +X/2)/2] + (Y +X/2)/2 = 2020

Now, Let's take the RIGHT part in (3) and equalize it with the LEFT part in (1)

0.5*(Y + X/2) = X

(2020-X) + X/2 = 2X

4040 - 2X + X = 4X

4040 = 5X

X = 808

So, Y = 1212

proof: 1) 808 + 1212 = 2020

2) 404 + 1616 = 2020

3) 1212 + 808 = 2020 #