Are you interested in math and puzzles? Do you like winning money? Well, why not participate in ThePuzzlr Contest, a 40-minute, 30 questions test for all ages that will be available to take via our website! We even have a sample test to help you get a feel of the problems! If you have already signed up, be sure to share this with as many friends as you can.

Puzzler Quik

There are 250 participants in ThePuzzlr completion. 200 of the participants are from the U.S.A. 100 of the participants like puzzles more than math! At minimum, how many participants like puzzles more than math and are from the U.S.A?

Puzzled?

Here is a hint to help you get going in the right direction:

Think about the best case scenario!

Hint 2:

There is addition and subtraction involved.

Puzzler Think

Bag A has 3 white and 2 black marbles. Bag B has 4 white and 3 black marbles.

Suppose we draw a marble at random from Bag A and put it in Bag B. What is the expected number of white marbles in Bag B

Puzzled?

Here is a hint to help you get going in the right direction:

Try making equations

Hint 2:

You don’t need a second hint!

ANSWERS TO LAST WEEK’S PUZZLERS

Last Week's Puzzler Quik

Yujen from Lancaster, Pennsylvania was the winner of this shout-out. He had the fastest entry time!

Let $x$ be the total amount of money. We then know that $\frac{x}{2}+\frac{x}{4}+\frac{x}{6}+1000=x.$ Combining like fractions, we get $\frac{11x}{12}+1000=x.$ We can then isolate x to get $x-\frac{11x}{12}=\frac{x}{12}=1000.$ Multiplying both sides by 12, we get that $x=12,000.$ Thus, Grandma had a total of $12,000.

Last Week's Puzzler Think

This week’s shoutout goes to Julia from New Jersey! Kudos to her for solving this problem!

Here is a way to do this problem:

Let $a_n$ be the number of colorings of an $n$ -gon with the desired property (where a “ $2$ -gon” is just two vertices considered adjacent to each other).

If $n=2$ , then we have three choices for the color of the first vertex and two choices for the color of the second vertex. Thus $a_2=3\cdot 2=6$ .

If $n=3$ , then we must make one vertex red, one green, and one blue, but we can assign these colors in any order. This gives us $a_3=3!=6$ .

If $n\ge 4$ , then we consider two cases. Either vertices $1$ and $n-1$ are different colors, or they are the same color.

First case: If vertices $1$ and $n-1$ are different colors, then vertices $1$ through $n-1$ are colored in a way which would be legal for an $(n-1)$ -gon, and the color of vertex $n$ is forced. Thus, within this case, we have $a_{n-1}$ legal colorings of the $n$ -gon.

Second case: If vertices $1$ and $n-1$ are the same color, then vertices $1$ through $n-2$ are colored in a way which would be legal for an $(n-2)$ -gon, and vertex $n$ can be either of two colors. Thus, within this case, we have $2a_{n-2}$ legal colorings of the $n$ -gon.

The two cases taken are exclusive and cover all possibilities, so we obtain the recurrence
$a_n = a_{n-1} + 2a_{n-2}\quad\text{for }n\ge 4.$
Making a table, we obtain:
$\begin{array}{c || c *6{| c}} n & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline a_n & 6 & 6 & 18 & 30 & 66 & 126 & \boxed{258} \end{array}$