In going through some folders in my office the other day, I came across some sets of mathematics puzzles I wrote for a conference of the International Group for Mathematical Creativity and Giftedness in 2014. Teachers of mathematics of all levels attended, from elementary school to university. The organizing committee (which included me) thought it might be fun to have some mathematical activity that conference attendees could participate in.

So I and my colleagues created three levels of contests — Beginning, Intermediate, and Advanced — since it seemed that it would be difficult to create a single contest that everyone could enjoy. But I did include three problems that were the same at *every* level, so all participants could talk about some aspect of the contests with each other.

Participants had a few days to get as many answers as they could, and we even had books for prizes! Many remarked how much they enjoyed working out these puzzles.

Now this conference took place before I started writing my blog. I have written several similar contests over the years for various audiences, and so I thought it would be nice to share some of my favorite puzzles from the contests with you. And so *The Puzzle Archives* are born!

First, I’ll share the three puzzles common to all three contests. I needed to create some puzzles which were fun, and didn’t require any specialized mathematical knowledge. As I’m a fan of cryptarithms and the conference took place in Denver, I created the following puzzle. Here, no letter stands for the digit “0.”

For the next puzzle, all you need to do is complete the magic square using the even numbers from 2 to 32. Each row, column, and diagonal should add up to the same number. There are two solutions to this puzzle — and so you need to find them both!

And of course, I had to include one of my favorite types of puzzles, a CrossNumber puzzle. Remember, no entry in a CrossNumber puzzle can begin with “0.”

I also included a few geometry problems, staples of *any* math contest. For the first one, you need to find the area of the *smallest* circle you could fit the following figure into. Both triangles are equilateral; the smaller has side length 1 and the larger has side length 2.

And for the second one, you need to find the radius of the larger circle. You are given that the smaller circle has a diameter of 2 units, and the sides of the square are 2 units long. Moreover, the smaller circle is tangent to the square at the midpoint of its top edge, and is also tangent to the larger circle.

The last two problems I’ll share from this contest are number puzzles. The first is a word problem, which I’ll include verbatim from the contest itself.

Tom and Jerry each have a bag of marbles. Tom says, “Hey, Jerry. I have four different colors of marbles in my bag. And the number of each is a different perfect square!” Jerry says, “Wow, Tom! I have four different colors of marbles, too, but the number of each of mine is a different perfect

cube!”If Tom and Jerry have the

sametotal number of marbles, what is the least number of marbles they can have?

And finally, another cryptarithm, but with a twist. In the following multiplication problem, *F,* *I,* *N,* and *D* represent different digits, and the *x*‘s can represent *any* digit. Your job is to find the number *F I N D. *(And yes, you have enough information to solve the puzzle!)

Happy solving! You can read more to see the solutions; I didn’t want to just put them at the bottom in case you accidentally saw any answers. I hope you enjoy this new thread!

(Note: The *FIND* puzzle was from a collection of problems shared by a colleague. The first geometry problem may have come from elsewhere, but after four years, I can’t quite remember….)

The simplest entry point into solving the first cryptarithm is to look at the fifth column. D would have to be 0 or 9, but it can’t be 0 since we are given there are no zeroes in the puzzle. The final solution:

For the magic square puzzle, you need to first add the even numbers from 2 to 32, which is 272. Since the four rows use all 16 numbers, each row, column, and diagonal sum to 272/4, or 68. The two solutions (which differ only by interchanging the middle two columns):

The solution to the CrossNumber puzzle:

Here is a diagram illustrating the solution to the first geometry puzzle. The essential idea is to use the geometry of the equilateral triangles to show that *ABC* is a right triangle, meaning that *AC* is a diameter of the circle. Since *AC* is then the area of the circle is

The second geometry problem is a little trickier. The diagram below will help, where *q* is the center of the larger circle. If we say the length of *pq* is 2 + *x,* then so is the length of *qr*, since they are both radii of the circle. Therefore, the length of *rs* must be 2 – *x.* Since the length of *sq* is 1, we may apply the Pythagorean theorem to triangle *qsr,* which gives *x* to be 1/8, so that the radius of the larger circle is 17/8.

As for marbles, Tom has 9 + 16 + 36 + 100 = 161 marbles, and Jerry has 1 + 8 + 27 + 125 = 161 marbles. Here, begin by looking for the smallest possible sum of four cubes (which won’t work); the next smallest sum is the one you want.

And for the final cryptarithm, start by observing that *N* must be 0, and because each partial product is only four digits long, this means that *F* must be fairly small. The answer is *FIND* = 3201.

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