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A Mathematical Zoo: A Wild Puzzle Compilation

Updated: May 17, 2022

Last month, we looked at one of my favorite investigative problems—involving coconuts, pirates, and of course, thieving monkeys—diving into the jungle to uncover fascinating arithmetic that can be traced back to the ancient Greeks. For July’s in-depth monthly edition, I bring you a companion article of sorts: I'll pose several more animal-themed puzzles to you to try. You won't want to miss the thrill of the chase on these wild puzzles, so go get some scrap paper and a pencil, and enter the zoo once again with me!

To add a little bit about my inspiration for this issue, I've noticed how much you all loved my previous puzzle compilation articles, so it seemed like the perfect time to bring that back. Here are a few previous compilations if you’d like to check them out:

Just like my previous puzzle compilations, I’ll place the puzzles in the main section of the article below and you can find solutions at the end. Note that some of the solutions are entirely complete, but others may guide you to find the answer yourself; in each case, I chose the approach that prioritizes the most enlightening process of discovery so you can get the most out of this article. Some of these puzzles are fairly easy on the surface but are actually much more complex, so make sure to read the in-depth solutions to get the most out of the process!

As you might imagine, animals are rather a broad and popular topic across the board in the math world, so it was quite difficult to narrow this article down to just these puzzles. I worked hard to make sure the puzzles I decided to include are accessible to practically everyone. There are so many amazing questions out there, so if you find any others that are particularly exciting, you are always welcome to share them with me and GLeaM in the forum.

1) Cats vs Rats

Source = Entertaining Mathematical Puzzles by Martin Gardner

Our first puzzle will serve as a warm-up. It’s unclear exactly where this puzzle originated because it’s been quoted so many times across the internet; it turns out it is even used frequently as a job interview question. It is, however, featured in one of Martin Gardner’s books, so that is the most-agreed upon source. Regardless, it’s a great example of when jumping to conclusions can be misleading.

If three cats can catch three rats given three minutes, how many cats would catch a hundred rats given a hundred minutes?

Here’s a bit more advice: The answer might not be the first one that comes to mind, but it also might not strictly be the second one either… you will want to think about proportions, but you will also want to think about the way cats catch rats in the first place. Is there really a good answer at all?

When you have an answer, go ahead and scroll to the end of the article!

My cats, Jasper and Coco, who would catch exactly zero rats even given a hundred years.

2) Jumping Frogs

Source = So You Think You’ve Got Problems? by Alex Bellos

This is yet another problem that has much more to it than initially meets the eye.

Fred the frog is sitting on the first lily pad in a line of lily pads. He wants to reach the tenth lily pad, and he can either jump to an adjacent lily pad (one down the line) or jump over an adjacent lily pad and land on the one after it (two down the line). If he never moves backwards, in how many ways can Fred reach the tenth lily pad?

If you’ve been in the math world for some time, it’s quite likely that you’ve seen many problems involving frogs jumping on lily pads. It is, after all, one of the easiest ways to set up a sequence or coordinate geometry problem in a more interesting and exciting way.* If so, this one may seem simple in comparison, but we’ll be able to take it much further than you might initially realize.

As you’re solving the problem, ask yourself the following further questions: What answers would we get if Fred wanted to reach the eighth lily pad? The ninth? The eleventh? The twelfth? Is there any pattern you can find? Could you find an expression for the answer if Fred wanted to reach the nth lily pad?

We’ll discuss this more in the solutions, but you can actually find the answer to Fred’s problem in two ways: recursively and explicitly using combinations aka choose functions. By setting these two solutions equal, we are going to prove a rather fun fundamental identity in combinatorics—all from a silly little lily pad problem! (If you’re ambitious, feel free to see if you can figure out what I mean before skipping to the solutions.) This is my favorite solution discussion, so make sure you read it!

Fred resting before his journey. Courtesy of Adobe Stock.

*If you don’t believe me, here are just a couple recent AMC problems involving frogs. If you want to solve any of them, simply click on the title of the problem and it will take you to the solution.

3) Ten Canaries

Source = So You Think You’ve Got Problems? by Alex Bellos

Originally phrased in terms of rats, this is one of my favorite puzzles to grasp the sheer concept of scale—and (hint!) the power of powers. The following is very close to a verbatim restatement of the puzzle, so all credit goes to Alex Bellos.

You have inherited a collection of one thousand bottles. All the bottles contain wine except one, which contains poison. The only way to discover what’s in a bottle is to drink it. If you drink poison, however, you die.

Thankfully, you have 10 canaries. If a canary sips any amount of poison, or poison mixed with wine, it will die after exactly one hour. If a canary drinks just wine, it survives. How do you determine which bottle is poisoned exactly one hour after the first canaries are given their first sips?

You should be able to come up with many schemes given unlimited time, but you only have one hour. What should you do?

Here is a hint: You are allowed to mix small amounts of any number of bottles together and give each canary a mixed concoction to sip from at the same time. That way, you will be able to gain information from the dead canaries at exactly one hour; all the dead canaries will have sipped from a concoction that contains the one poisoned bottle. However, you will only have ten pieces of information. Which bottles should you have each canary sip from to identify the exact poisoned bottle out of a thousand bottles with only ten canaries? What scheme should you follow?

Two canaries. Courtesy of BioMed Central.

4) Fox and Goose

Source = Entertaining Mathematical Puzzles by Martin Gardner

Courtesy of Martin Gardner’s Entertaining Mathematical Games

Fox and Geese is a classic hunting-themed board game that involves one player as the fox and other players as the geese. The geese attempt to surround and capture the fox so he cannot move in any direction, while the fox attempts to capture and “eat” geese until there are too few geese left to be able to trap him. For a more in-depth description of the rules of the game, click here.

In the above simplified version, we only have one goose, and because that means there are not enough geese for the geese to surround the fox and win the game in the traditional sense, we focus solely on how soon the fox can capture the goose. We dub this “Fox and Goose.”

Begin with the fox at 21 and the goose at 31. Each move involves the fox or the goose moving from one dot to an adjacent dot over one black line segment. The fox’s goal is to capture the goose by landing directly on the spot it occupies.

Can the fox always capture the goose in ten moves? Or can the goose always manage to get away in ten moves? What strategy must either the fox or the goose employ to ensure their win (in ten moves), if either has a dominant strategy?

Note: This puzzle comes from a branch of math called game theory, which is essentially the study of games and the strategies opponents employ to win. A dominant strategy is a series of moves one player is able to use to win no matter what the other player attempts to do to stop them. Not every game has a dominant strategy for one of the players, but some do, and there is certainly a lot of fascinating math in looking for these strategies and examining various games. If you’d like to learn more, click here.

Courtesy of the Fox and Goose Hotel in London.

5) Thirteen Camels

Source = So You Think You’ve Got Problems by Alex Bellos

This is as much a brain teaser as a math problem, and I think you’ll like it!

This puzzle has actually been in the popular historical canon for quite some time as a similar statement involving seventeen camels, and the following is a quoted excerpt from Alex Bellos’ adaptation.

A man leaves 13 camels to his three children in the following proportions: to the eldest, half; to the middle child, a third; to the youngest, a quarter. The children can’t decide how many each should get, because you cannot divide 13 by 2, 3, or 4 without inflicting severe pain on a camel. They consult a wise old woman to resolve their dispute. How does she fix it?

Before you skip to the solutions, make sure you read the problem very carefully. Does the man’s will even make sense in the first place? Is there a way to maintain the essence of the will without creating fractional camels? This puzzle isn’t entirely objective, so this will require thinking outside of the box. With a little thought, you should be able to decide what the best answer is for our children—and find a way to preserve what is most important.

Courtesy of Pexel.


As I was researching for this article, I found many more puzzles than I could possibly include. If you still are looking for more questions, here are a few more I love that didn’t quite make the cut. I am not including solutions for these, but you are more than welcome to try them and let me know what you think!

For whatever reason, it seems my animal puzzles lean more heavily towards number theory and combinatorics—perhaps because these types of problems lend themselves more easily to word problems? If you find any geometry-based, algebra-based, or other alternative questions, I would be happy to take suggestions to add to this section!

  1. There is a large cube formed by gluing together 27 smaller cubes of uniform size. A termite starts at the center of a face of any of the outside cubes and bores a path that takes him once through each cube. His movement is always parallel to a side of the large cube, never diagonal. Is it possible for the termite to bore through each of the 26 outside cubes once and only once, then complete his journey by entering the central cube for the first time? If possible, show how; if not, prove it. (Source: Martin Gardner’s Colossal Book of Short Puzzles and Problems)

  2. A dealer bought a number of horses at $344.00 each, and a number of bullocks at $265.00 each. He then discovered that the horses had cost him in all $33.00 more than the bullocks. What is the smallest number of each that he must have bought? (Source: Henry Ernest Dudeney’s 536 Puzzles & Curious Problems)

  3. A colony of chameleons on an island currently comprises 13 green, 15 blue, and 17 red individuals. When two chameleons of different colors meet, they both change their colors to the third color. Is it possible that all chameleons in the colony eventually have the same color? (Source: Alex Bellos’ So You Think You’ve Got Problems?)

  4. A fly lands on the inside of a cylindrical glass tumbler, 2 cm from the rim. A spider is on the opposite of the glass, 2 cm from the bottom. The glass has a height of 8 cm and a circumference of 12 cm. If the fly remains stationary, what is the shortest distance the spider must crawl in order to reach it? (Source: Alex Bellos’ So You Think You’ve Got Problems?)

  5. Our diagram represents a field 100 yards square. Pat and the pig that he wishes to catch are in opposite corners, 100 yards apart. The pig runs straight for the gateway in the top left-hand corner. As the Irishman can run just twice as the pig, you would expect that he would make straight for the gate and close it. But this is not Pat’s way of doing things. He goes directly for the pig all the time, thus taking a curved course. Does the pig escape, or does Pat catch it? And if he catches it, exactly how far does the pig run? (Source: Henry Ernest Dudeney’s 536 Puzzles & Curious Problems)

Courtesy of Henry Ernest Dudeney; Markings by me.

And… my favorite :) from the 2019 UGA High School Math Tournament:

For n > 1, a geometric figure is called an n-reptile if it can be decomposed into n figures, which are congruent to each other and similar to the original figure. For example, a square is a 4-reptile and also a 9-reptile.

  1. What is the smallest n > 1 for which a 30-60-90 triangle is an n-reptile?

  2. For how many n, 1 < n < 50, is an equilateral triangle an n-reptile?

  3. If a right triangle with shortest leg 1 is a 5-reticle, what is the length of the hypotenuse?

Get it? A rep-tile (i.e. a repetitive tile) can also a reptile, as in a lizard or a snake!

A green iguana. Courtesy of National Geographic.

Rep-tiles are a super fun concept to play with, so click here if you want to learn more!


1) SOLUTION: Cats vs Rats

The usual solution to this puzzle goes like this: If three cats can catch three rats in three minutes, that means the three cats (as a team) catch one rat in one minute. This translates to the same three cats catching a hundred rats in a hundred minutes. Thus, the answer is three cats, not 100 cats as many people tend to blurt out.

As an interview question and a riddle, the question tests your ability to think critically about whether the number of cats needs to change—after all, both the number of rats and the number of minutes increased simultaneously, not just one or the other, and the effect essentially cancels itself out—and to not fall blindly to a pattern that seems to indicate 100 is the answer.

This puzzle is already clever enough to warrant an addition to this article, but there is more. Remember that I said the second answer you think of (which was, hopefully, 3) might not strictly be the answer either? I also asked you to think about the way cats catch rats to begin with. Do you see any potential problems with an answer of 3?

As Martin Gardner points out in his book, we actually have zero reason to assume that three cats taking three minutes to catch three rats also implies that the three cats take one minute to catch one rat. We’re assuming that the three cats work together as a team, putting all their energy into catching one rat at a time each minute—when it is actually quite possible that each cat works independently, catching their own rat every three minutes.

Because we simply do not know the process the three cats follow in the three minutes to catch the three rats and thus, we cannot assume that they catch rats at a constant rate as a group (a fact we need to be able to work with proportions in this way), we only know for certain how long it takes the cats to catch numbers of rats that are already multiples of 3.

(Here, we are only really assuming that the cats keep up their combined rate throughout the entire time they are catching rats, which is a more justifiable assumption. To add even more intricacies—such as claiming they will get tired and slow down over time—would quickly become an exercise in splitting hairs and probably require a lot more serious mathematical modeling.)

From here, we know the cats will take 99 minutes to catch 99 rats, but we have no idea how long it will take them to catch the last rat. If they work independently, one cat will claim the rat themselves and take a full three minutes, meaning it will take them 102 minutes to catch 100 rats.

In short, the question is simply too ambiguous for us to make a proper conclusion. We might need a fourth cat to pick up the slack (since we can’t have fractional cats), but we might not. Due to the lack of information, we’re unfortunately left between a rock and hard place—with no satisfying answer to our problem. See how there’s often much more to a situation than you first realize? It’s a technicality, but it is certainly one we should not have originally overlooked.

2) SOLUTION: Jumping Frogs

The answer is 55, the ninth Fibonacci number. There are several methods to solve this problem, but we will start with the recursive one. (Simply put, ’recursive’ means that we are not finding the solution through direct calculation or formulas, but instead by looking at the answer for smaller values of lily pads and building up our answer successively using arithmetic and straightforward rules about our situation. For more, click here.)

Essentially, all we are doing is simplifying the problem and finding the pattern. Let’s start with small cases.

What if Fred’s goal was to reach the second lily pad? In this case, there is only 1 way to get to the last lily pad: one jump of length 1.

What if he wanted to reach the third? Now, there are 2 ways to reach the last lily pad: two jumps of length 1 or one jump of length 2.

The fourth? There are 3 ways: a jump of length 1 followed by a jump of length 2; a jump of length 2 followed by a jump of length 1; or three jumps of length 1.

What about the fifth? There are 5 ways: two jumps of length 2; a jump of length 2 followed by two jumps of length 1; a jump of length 1 followed by a jump of length 2 followed by a jump of length 1; two jumps of length 1 followed by a jump of length 2; or four jumps of length 1.

We can continue this way, merely counting all the sequences of 1’s and 2’s that sum up to our desired total jump length. (Notice that we need the total jump length to be one less than the desired lily pad number because it only takes one jump to get from the first to the second lily pad and similarly, it only takes four to get from the first to the fifth.) However, there is a much more clever way to look at it.

Let’s think about the five lily pad case again—and let’s consider only Fred’s first jump. His first jump is either length 2, in which case there will be three lily pads left, or it is length 1, in which case there will be four lily pads left. Therefore, the total number of cases is the number of ways Fred can jump to the third lily pad (2) plus the number of ways he can jump to the fourth lily pad (3): 2 + 3 = 5.

Similarly, if Fred wants to reach the sixth lily pad, we sum our solutions for four and five lily pads: 3 + 5 = 8. And, for the seventh, we sum our solutions for five and sixth lily pads: 5 + 8 = 13. Then, for the eighth, we have 8 + 13 = 21; for the ninth, we have 13 + 21 = 34; and for the tenth, we have 21 + 34 = 55. Thus, 55 is our answer.

This forms the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55…

If you’ve seen it before, the Fibonacci sequence is simply the sequence where each term is formed by adding the two previous terms. It is also largely considered the most famous recursive sequence. The first 1 is generally considered the zeroth term, so when solving for the nth lily pad, we must adjust slightly, and we have that our answer is the (n-1)th Fibonacci number, often written F_(n-1).

There is an explicit formula for the Fibonacci sequence, but it is rather complicated, so this is enough of a closed form for us at the moment. If you’d like to read even more about the Fibonacci sequence, feel free to check out one of my favorite articles from last year here.

What is so neat about this puzzle is that it naturally generates the Fibonacci sequence because the frog’s options to proceed forward are naturally based on the sum of the options for its two previous destinations, giving us a rather familiar concept in a seemingly unrelated scenario.

Wait there’s more! We may have already solved the problem, but I promised you a second method—and some deeper mathematical truth.

We can also attempt to calculate the problem directly as a series of cases based on the number of jumps of length 2 the frog takes. After all, the problem also boils down to counting sequences of 1’s and 2’s, and the second approach relies on this.

If Fred wants to reach the tenth lily pad, he can choose from the following options:

  • Case 1: 4 jumps of length 2, 1 jump of length 1

  • Case 2: 3 jumps of length 2, 3 jumps of length 1

  • Case 3: 2 jumps of length 2, 5 jumps of length 1

  • Case 4: 1 jump of length 2, 7 jumps of length 1

  • Case 5: 9 jumps of length 1

Of course, each of these cases (except for the last) has several ways to complete it—based on the order in which Fred chooses to use his jumps. This implies that we can use combinations (click here for more).

Here are the calculations:

As you see, this also gives us the correct answer, 55, the ninth Fibonacci number. This is a relatively straightforward and remarkable method in and of itself, but you’ll notice that we can also apply this idea to any number of lily pads.

That means that if this series gives us a Fibonacci number, any corresponding series should also give us its corresponding Fibonacci number, which gives us an identity for Fibonacci numbers as sums of combinations.

Here is the expanded version:

Courtesy of Art of Problem Solving.

Look how incredible it is that this problem held such a beautiful equation within it!

When solving a problem, it is always important to tie your loose ends. The elegance of the equation for ten lily pads and our easily-expandable logic surely implies it will hold for all values of n, but that is not a full proof.

Therefore, my only remaining challenge for you is this: Convince yourself that this equation is indeed equivalent to solving Fred’s problem two ways. Prove completely that our equation is true, using Fred as a guide. When you have a proof, feel free to email it to me ( for feedback and advice.

3) SOLUTION: Ten Canaries

The essence of this puzzle lies in the binary number system and the scale of powers of two.

As I explained above, our strategy is to mix small amounts of a large number of bottles to make ten different concoctions for our ten canaries to drink simultaneously. Then, after one hour, we will hopefully be able to use the ten pieces of information of whether each canary is alive or dead to determine which one of the thousand bottles of wine was poisoned.

Let’s work backwards: We notice that the information we obtain at the end of the hour is whether each canary is dead or alive. For example, we might have the following list:

  • Canary 1 = Dead

  • Canary 2 = Dead

  • Canary 3 = Alive

  • Canary 4 = Alive

  • Canary 5 = Dead

  • Canary 6 = Alive

  • Canary 7 = Dead

  • Canary 8 = Alive

  • Canary 9 = Dead

  • Canary 10 = Dead

Instead of using the words “Dead” or “Alive,” we can also choose something equivalent to encode our information—like digits. If we allow “Dead” to correspond to the digit one and “Alive” to correspond to the digit zero, we can act like computers and line the canaries up from Canary 10 to Canary 1 and form a number that encodes all of our information quickly: 1101010011.

Wait a second… that looks like a binary number (click here for more)! And despite our initial impression that we might have too little information, we actually have 2^10 = 1024 possibilities for our end result—slightly more than we need to differentiate each of our one thousand potentially poisoned bottles from the other 999.

Now, we just need a scheme that forces all the correct canaries to die so that our end result binary number (as formed through dead or alive canaries above) is exactly the same as the numbered wine bottle (1 to 1000, base 10 OR 1 to 1111101000 in binary) that contains the poison.

Here is that scheme: We assign each canary to a place value in our binary number.

Canary 1 is the units, Canary 2 is the twos, Canary 3 is the fours, and so on.

We then give each canary a portion of each bottle that has a “1” for their corresponding place value. That way, we create the exact scenario that we engineered backwards earlier: the canary will die if the poisoned bottle has a “1” in the place value that corresponds to that canary. Thus, the digit 1 acts exactly like the information given by the “Dead” canary—and similarly, the digit 0 acts like an “Alive” canary.

That’s it! Lining our canaries up from Canary 10 back to Canary 1, we can form the exact binary number we need and trace it back to our poisoned bottle. Thus, we can successfully get rid of that bottle and form a complete unpoisoned collection!

I have one last question for you: Say you want to keep two canaries as pets after you are done testing your wine. Can you come up with a scheme that ensures that at least two canaries will live?

4) SOLUTION: Fox and Goose

The fox can always capture the goose in the ten moves allotted. In the first three moves, she moves around one of the central triangles in the middle of the board, at which point the goose has moved to one of the following spots: 13, 20, 25, 27, 32, 34. (See if you can convince yourself that this is a complete list.) At this point, the fox is in a position to be able to trap the goose back into its corner and eat it.

Martin Gardner includes this example of a game played with his dominant strategy in his book:

Courtesy of Martin Gardner’s Entertaining Mathematical Games

This is not quite a full explanation, so I will hand the rest to you. Can you write a full description of the strategy the fox must employ to capture the goose no matter what series of moves the goose follows? How can you use the series of possible spots for the goose after the first three moves that we found above? Is the required number of moves for the fox to win ten or is it even less?

What is the significance of the triangle in the middle of the board—why is it important that the fox can get back to spot 21 in three moves at the beginning (rather than two or four)? Is the triangle absolutely critical to the fox’s strategy? Can you write a scheme that doesn’t include it?

I love this problem because it’s simple on the surface, but it is also a gateway to a much more in-depth investigation, much like the one we did last month. If you answer any of the questions above, feel free to discuss them in the comment section, the forum, or by emailing me; I would love to have a discussion with you!

Hint: Of course, what makes a game theory investigation even more fun is that you can play these puzzles, so grab a sibling or a friend, make a board, and see what you must do to ensure you always beat them.

In fact, when you’re done with this board, you can expand the question even more. What happens if you play on larger boards—with or without additional triangles or diagonal lines? Could you even make this game 3D? What are the minimum number of moves required for the fox to always win in these expanded scenarios? Can the fox always win at all? Feel free to ask your own questions, and I’d love to hear what you can come up with.

5) SOLUTION: Thirteen Camels

Just like our first puzzle, the thirteen camels are a lesson in not jumping to conclusions—and looking for ways numbers can lead us astray. The problem with the father’s will was not only the fact that it implied fractional camels (13/2, 13/3, and 13/4 are certainly not integers), but also that his proportions did not add up to 1 anyway. It is easy to overlook the latter problem while focusing on the former, but of course, this is an even deeper issue. Even given a material that can be divided into exact proportions (like, say, a barrel of ancestral orange juice), there would be no way to give the oldest one half, the middle child one third, and the youngest one fourth. 1/2 + 1/3 + 1/4 equals 13/12 not 1, so there would simply not be enough to go around.

How does the wise woman resolve this? She temporarily steals one of the camels, telling the children that it escaped in the dead of night and they will not see it again.

With 12 camels remaining, she now has easy integer proportions to give the children according to their father’s will, handing out 12/2 = 6 camels to the oldest and 12/3 = 4 camels to the middle child. She then gives the 2 remaining camels to the youngest alongside the one she hid earlier, for a total of 12/4 = 3 camels. Of course, none of the children know the stolen camel is actually alive, so they all believe they’ve received exactly what they were owed.

In this solution, not only do the children believe they received the exact proper proportion of (whole!) camels, but the wise woman also does not break any laws of the land: she skillfully distributes the entirety of the man’s property to his children in the most equitable way possible.

In Alex Bellos’ solution, the woman confiscates the camel in full daylight and in agreement of the children and convinces them she has the best possible outcome… but I think theft is a bit more fun :)

In short, because our father’s will didn’t create possible fractions in the first place, we can maintain the essence of the will by giving the children camels in the proper proportions of 1/2: 1/3: 1/4 = 6/12: 4/12: 3/12, which is exactly what we did. Indeed, the father never states that each child should have a half, a third, and a quarter of the total 13 camels, but he instead emphasizes that the proportions will be correct—which can be justified as the most important aspect of his will. Thus, the wise woman followed his rules exactly.

In the original 17-camel puzzle, the man asks for his children to receive his 17 camels in the following proportions: one half to the eldest, one third to the middle child, and one ninth to the youngest. How do you think the wise woman would resolve the issue in this case?


Thank you so much for reading this month’s article! It was so much fun to compile all of these problems for you, and I hope you enjoyed it.

I highly recommend all of the books I used as sources in this article, especially Entertaining Mathematical Puzzles and The Colossal Book of Short Puzzles and Problems by Martin Gardner, 536 Puzzles & Curious Problems by Henry Ernest Dudeney, and So You Think You’ve Got Problems by Alex Bellos. If you are ever looking for more exciting problems to solve, these books have you covered, and I definitely suggest checking them out!

As always, I am more than willing to receive feedback, questions, and comments about anything in the comment section, in the forum, or in my inbox ( Until next time, enjoy puzzle-solving!

Cover Image: An M.C. Escher-inspired tessellation by Mak & Nak - NICOLAS

Note: GLeaM will be temporarily suspended for August while I make the transition to college. However, there may be more content in the near future, and all members and subscribers will get an email notification if and when the next edition is published.



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