Math Puzzles About Donkeys
Donkeys are the underclass of the equine genus, often the butt of jokes and the inspiration for insults that humans lob at other humans, most notably "ass." Donkeys are typically associated with poor people and the underdeveloped world, while horses are associated with rich people and sophistication.
Though they are not as beautiful and strong as horses, and not as easy to train, donkeys still have their quirky charms. They are not as easily spooked as horses and have remarkable endurance. Studies show they are smarter than horses too. Humans have used donkeys as pack and draft animals for over 5000 years, and in that time span donkeys have proven to be loyal and smart companions. To honor these oft-maligned creatures, here are several math puzzles and brain teasers about donkeys.
(1) Queen of the Hill
Kathy is a donkey who likes to graze at the top of a hill. She won't let any other donkeys graze with her unless they agree that she is the donkey queen and they give her a challenging riddle to solve. One day four donkeys named Al, Bill, Cal, and Dell come to the hill and ask Kathy if they can graze with her. Al says:
"My queen, yesterday Bill, Cal, Dell and I carried 48 sacks for farmer Paul. Dell carried three more sacks than I, I carried one more sack than Cal, Cal carried two more sacks than Bill, Bill carried three more sacks than Dell, and one of these statements is a lie. Can you determine which statement is a lie and how many sacks each of us carried?
Kathy thinks about it for a few minutes, solves the riddle, and then tells the donkeys to get off her hill unless they come up with a better puzzle. Can you solve the donkeys' riddle?
Solution: The statement "Cal carried two more sacks than Bill" is a lie. Of the 48 sacks, Bill carried 16, Dell carried 13, Al carried 10, and Cal carried 9.
To solve this problem we can translate the clues into algebra:
- A + B + C + D = 48
- D = A + 3
- A = C + 1
- C = B + 2
- B = D + 3
One of the last four must be a lie since they form a contradictory cycle. We can proceed by elimination.
- If "D = A + 3" is a lie, then the other four statements imply that 4D + 14 = 48, or D = 8.5. But non-integer answers do not make sense in the context of the problem. One sack is one sack whether it is a small sack or large sack, partially full or not. Thus, we can discard this possibility.
- If "A = C + 1" is a lie, then the other four statements imply 4A + 17 = 48, which gives A = 7.75. Again, this can be discarded because it is not an integer.
- If "C = B + 2" is a lie, then the other four statements imply 4C + 12 = 48, which gives C = 9. This answer makes sense.
- If "B = D + 3" is a lie, then the other four statements imply 4B + 11 = 48, which gives B = 9.25, again a nonsensical answer.
Therefore, we have
- C = 9
- A = 9 + 1 = 10
- D = 10 + 3 = 13
- B = 13 + 3 = 16
(2) Scratching Asses
Will and Dandy are two donkeys who love scratching their butts on fences. Their owner, farmer Paul, keeps them in a circular pasture 200 yards in diameter surrounded by a fence with plenty of scratching surface. Will and Dandy start scratching at the same spot due north of the enclosure, then they move around the enclosure with Will moving left and Dandy moving right, both donkeys scratching at every spot along their way. Will moves around the perimeter faster than Dandy, and after a while they find themselves scratching their butts on a common spot 150 yards from where they started. How much faster is Will than Dandy?
Solution: The point where they start, the point due south of that, and the point where they meet form a right triangle with a hypotenuse of 200, one leg length of 150, and another leg length of approximately 132.22876 using the Pythagorean Theorem.
In the diagram above on the left, angle θ is equal to arccos(150/200) ≈ 41.40962 degrees. In the diagram above on the right, angle φ is approximately equal to 180 - 2(41.40962) = 97.18076 degrees. The proportion of the circumference traversed by Dandy the donkey is 97.18076/360, or about 0.26995 . The proportion traversed by Will is (360 - 97.18076)/360 ≈ 0.73005. The ratio of Will's speed to Dandy's is about 0.73005/0.26995 ≈ 2.70444. In practical terms, Will is about 2.7 times faster than Dandy when it comes to fence butt scratching.
(3) Donkey Chat
Morris the donkey prefers to chat with himself even when there are other animals around to talk to. Because he is a math-minded donkey, he talks to himself in a peculiar pattern. Every new sentence he says to himself has half as many words as the previous sentence if the sentence has an even number of words. Otherwise, it has one more word than the previous sentence if the previous sentence has an odd number of words.
What happens when Morris's first sentence has 13 words? What eventually happens to the pattern of Morris's speech when his first sentence has N words?
Solution: If Morris's first sentence has 13 words, then the number of words in the subsequent sentences will be
14, 7, 8, 4, 2, 1, 2, 1, 2, 1, ...
It devolves into a cycle of alternating one-word and two-word sentences. If Morris's first sentence has 19000 words, the number of words in the subsequent sentences will be
8500, 4250, 2125, 2126, 1063, 1064, 532, 266, 133, 134, 67, 68, 34, 17, 18, 9, 10, 5, 6, 3, 4, 2, 1, 2, 1, 2, 1, ...
As you can see, this also devolves to the same pattern. It turns out that no matter how long his first sentence is, eventually he will reach the 2, 1, 2, 1,... pattern. You can understand this more intuitively by looking at the two opposing forces -- growth and decay -- that act on the numbers. The force of decay -- halving -- is much stronger than the force of growth -- adding 1 -- for numbers larger than 2. Moreover, the decay force acts at least as often as every other turn (because adding 1 to an odd number produces an even number), so overall the sequence decreases until you can only oscillate between 2 and 1.
Aside: This problem is superficially similar to an unsolved conjecture in number theory called the Collatz Conjecture. Suppose when the previous sentence has an odd number of words N, the number of words in the next sentence is 3N + 1, rather than N + 1. In other words, Morris halves or triples plus one. It is conjectured that no matter what number you start with this sequence always reaches the cycle 4, 2, 1, 4, 2, 1, ... However, nobody has yet found a rigorous proof of this fact. With computers, mathematicians have shown the conjecture is true for all positive starting values up to 2^60 or roughly 10^18.
(4) Donkey on a High Horse
Marita the donkey thinks she is a horse. The geese who live on the farm cannot tell the difference between braying and neighing and believe Marita when she says she is a horse. The dogs of the farm are divided; 60% of them know she is a donkey and the other 40% think she is a horse. All of the horses know Marita is a donkey, but sometimes they like to have fun at her expense by all pretending they think she is a horse too.
One day, Marita brays in front of an audience of 100 assorted geese, dogs, and horses; 88% of the animals express that she is a horse. The next day, she brays in front of the same audience, but only 42% of them express that she is a horse. How many horses, dogs, and geese are in the audience?
Solution: On the first day the horses are pretending Marita is a horse and on the second day they are being honest. Since only the horses change their opinion, we know that 88% - 42% = 46% of the animals are horses. Since there are 100 total, this means there are 46 horses.
On the first day, 100 - 88 = 12 animals express that Marita is a donkey, and these 12 are all dogs. These 12 represent 60% of the dog total. This means there are 20 dogs in all.
Finally there must be 100 - 46 - 20 = 34 geese.
(5) Donkey Lemonade
Rags the donkey pulls the beverage cart for the field laborers. Rags is mischievous and doesn't like humans; when she thinks nobody is looking, she drinks half the lemonade from a glass and replaces the missing liquid with donkey spit. She tries to do this for as many glasses of lemonade as she can.
The field laborers know what Rags is up to. Periodically when Rags isn't looking, the laborers switch around the glasses randomly so that Rags won't know which she has tampered with already, thereby increasing the chances that she drinks her own spit.
One day, Rags brings six glasses of lemonade on the beverage cart. She finds time to contaminate the glasses three times. Between spitting sessions, the workers switch around the glasses randomly. What are the chances that she ends up drinking her own spit at least once?
Solution: This problem is easier to solve by finding the probability that she doesn't drink her own spit. For the first glass, she has a 100% chance of getting a clean drink because they are all lemonade. For the second glass, her chances of getting a clean drink are only 5/6, because one of the glasses now has her spit. For the third glass, her chances are 4/6, because two glasses are now contaminated. The total probability that she doesn't drink her own spit is 1*(5/6)*(4/6) = 20/36 = 5/9, or about 55.6%. The probability that she does is 4/9, or about 44.4%.
(6) Don Quixote
Dawn Ki is a beautiful and acclaimed donkeyologist who holds advanced degrees from elite American universities, and who has written many best-selling novels about donkeys. Deranged horse-lover Tess Tea has a jealous obsession with Dawn and frequently writes angry 13-page letters to Dawn's publisher claiming Dawn is a dumb ugly donkey, and that they should publish Tess's horse novels instead. Along with her nasty letters she sometimes sends 260-page novel manuscripts about a horse named Tess who is prettier and smarter than everyone else.
The publisher forwards the correspondence to Dawn, who shreds the pages without reading them to make bedding for her donkey, Don Quixote. The donkey uses 1500 shredded pages a week. Last year, Dawn shredded 10 times as many letters as novel manuscripts to make bedding for her donkey. How many letters and novels did she shred?
Solution: If you assume 52 weeks in a year, then Dawn's donkey uses 52*1500 = 78000 pages in a year. Let L be the number of letters and M be the number of manuscripts. Two relations that relate L and M are
- 13*L + 260*M = 78000
- L = 10*M
If we replace L with 10*M in the first equation we get
130*M + 260*M = 78000
390*M = 78000
M = 200
This means that Dawn shreds 200 manuscripts and 2000 letters.
(7) The Meddlesome Mule
Grace is a mule, the sterile hybrid offspring of a horse and donkey. She is very proud of her horse heritage, but she has the obstinate disposition of a donkey. On the farm, Grace looks down on the donkeys because she thinks they have too many foals that they don't care for properly. She is always chastising them to be more like the horses, who have fewer foals but care for them better. She says to one of the donkey females:
"You have too many foals! If that horse over there took two of your foals you would have the same number. But if that horse gave you one of her foals, you would have four times as many as she!"
Before the donkey could give a snappy reply, Grace trotted away to pester another farm animal. How many foals does the donkey have, and how many does the horse have?
Solution: If the donkey has D offspring and the horse has H offspring, then Grace's statements translated into algebra are
- D - 2 = H + 2
- D + 1 = 4(H - 1)
The first statement is equivalent to D = H + 4, in other words, the donkey has four more foals than the horse. Plugging this into the second statement gives
(H + 4) + 1 = 4(H - 1)
H + 5 = 4H - 4
5 + 4 = 4H - H
9 = 3H
3 = H
Thus, the horse has three foals and the donkey has seven. This problem is a variation on Euclid's Ass and Mule Problem, a classic puzzle that was supposedly posed by the Greek mathematician Euclid. Puzzles of this type resemble math problems about people's ages and are solved the same way.
(8) The Donkey Doctor's Diagnosis
Clark the veterinarian informs farmer Paul that his favorite old donkey, Rando, has a bad case of talkativitis, a rare disease that causes donkeys to talk and get ideas about things. It is a highly contagious disease that will spread to all the other donkeys on the farm unless Rando is quarantined. Unfortunately, Rando always finds ways to get out of his pen.
Talkativitis spreads like this: An old donkey can infect exactly one old donkey and one young donkey. A young donkey can infect exactly one old donkey and two young donkeys. Once infected, a donkey will not be contagious until a full day has passed. How many days will it take for at least 900 donkeys (including patient zero) to have talkativitis? Assume on Day Zero only patient zero is infected.
Solution: Let's use the symbol "O" for an old donkey and "Y" for a young donkey. Starting with "O" on day zero, the next day we have "OY" or two infected donkeys. For the second day, we take the string "OY" and replace the "O" with "OY" and replace the "Y" with "OYY." This gives us "OYOYY" or five donkeys on the second day. Continuing this replacement algorithm over and over gives us each new day's list of infected donkeys.
- O (1)
- OY (2)
- OY OYY (5)
- OY OYY OY OYY OYY (13)
- OY OYY OY OYY OYY OY OYY OY OYY OYY OY OYY OYY (34)
As you might have noticed, the number of newly infected donkeys on each passing day is a Fibonacci number, every other Fibonacci number in fact. After four days have passed, the number of donkeys infected is 1 + 2 + 5 + 13 + 34 = 55. To get to 900, we just need to add up every other Fibonacci number until we reach a sum that is at least 900. It turns out that we need eight terms.
1 + 2 + 5 + 13 + 34 + 89 + 233 + 610 = 987
This means in seven days more than 900 donkeys will have talkativitis.
Photos via Pixabay and diagrams created by author.