Consider the value

				$|2^{4003}-  3^{2022}|$ 
			
		
	. Is this value at least $\\frac23$? Of course, you cannot use a calculator and have to explain how you get to your answer.

1. Simplify

First we have to determine what type of value  $|2^{4003}- 3^{2022}|$ is. It is a non-negative integer. To ensure it is at least $\frac23$, we simply make sure it is not zero.

2. Not Zero

If $2^{4003}= 3^{2022}$, then we know that (overkill) $2^{4003}= 3^{2022}\ \ (\text{mod 2})$ which is $0=1\ \ (\text{mod 2})$ which is not true.

Your favorite noodle place in the world, PACT China Noodle House, is known for its variety of different noodles. PACT China Noodle House has 3 different kinds of soup bases and 5 different kinds of noodle. In addition, you can choose to add at most one of any 10 different toppings (note: you can choose no topping as well). On top of all of these choices, you can choose spicy or non-spicy. How many different noodles can you possibly order from PACT China Noodle House?

Simple. 3X5X(10+1)X2=330 different noodles.

A bag has 2021 blue marbles and 2022 red marbles. We repeatedly remove 2 marbles from the bag. If the two chosen marbles are of the same color then we put 1 new red marble in the bag (after removing the 2 chosen marbles). If the two marbles are of different colors then we put one new blue marble in the bag. What will be the color of the last marble in the bag? Explain.

1. Recurse

Let function $f(r,b)$ return the result with $r$ red marbles and $b$ blue marbles.

$f(r,b) = f(r-2+1,b)$ or $f(r-1,b-1+1)$ or $f(r+1,b-2)$

2. Hey!

No matter how the recursion is done, the number of blue marbles remain the same parity.

Thus, the left over marble must be blue, as 2021 is an odd number.