Here, if we take the original number of pieces of paper as n. We can see that every time John tears the paper, it adds 2 pieces to the total count (n+2), while Sam adds 4 pieces to the count (n+4). Both of these additions do not change the parity of the number, which in this case, the total number of pieces of paper always stays as an odd number . Hence we can deduce it is impossible to have 100 pieces of paper.
Again if we had the same rules but the game has started off with 4 pieces of paper, would it be possible to have a total of 100 pieces of paper? Have a think about it.
Let’s have another example. If we take an 8x8 checkerboard, and we want to cover it with by a 1 x 2 domino, it is not hard to see that we can cover up the whole checkerboard full. If we take out the two corner squares, it is possible to cover up the remaining squares?
Use these to fill up the whole board!
With parity arguments, we can disprove this possibility. Since every adjacent square of a black square is a white square, the 1 x 2 domino must cover 1 white square and 1 black square. If we count up the number of black squares and white squares, we can see that there are a total of 32 black squares and 30 white squares. It is impossible to pair up all the black squares with enough white squares. Thus, it is impossible to fill up the whole area with 1 x 2 dominos.
Fun fact: The word parity comes from the pairing and equivalence of things.
There is also another type of parity: the parity between positive and negative numbers. When two negative numbers are multiplied, you get a positive number; when you multiply a positive and a negative, the result is a negative number; and lastly when you multiply two positive numbers, you get a positive numbers. This is common sense – but we can observe these positive and negative numbers follow a similar pattern as the odd-even parity. We can use these rules to solve the problem below:
18
Made with FlippingBook - professional solution for displaying marketing and sales documents online