Try solving ten medieval math riddles first posed by the scholar Alcuin of York. These puzzles from the early Middle Ages reveal how logic and learning were taught in Charlemagne’s time — and they’re still tricky to solve today.
By Lorris Chevalier
Albinus Flaccus Alcuin (735–804), a monk, scholar, and teacher, was among the most learned minds of the early Middle Ages. Appointed by King Charlemagne (c. 742–814) as tutor and adviser, he played a key role in reforming the educational system of the Carolingian Empire. In addition to his theological and pedagogical writings, Alcuin compiled a fascinating Latin collection entitled Propositiones ad acuendos juvenes — Problems to Sharpen the Wits of the Young.
This remarkable work presents 53 recreational puzzles: 33 on arithmetic (including eight dealing with division of goods), 12 on geometry, and eight on logic (four concerning river crossings). As noted by Pierre Dedron and Jean Itard in Mathématiques et mathématiciens, it represents “one of the earliest examples of mathematical recreations.”
Each problem is accompanied by its solution. Typically, Alcuin begins by stating the answer and then provides a brief justification. Of course, the mathematical tools and methods of his time did not allow for the kind of detailed reasoning we expect in modern solutions — yet his ingenuity remains striking even today.
Proposition 1 – Of a Snail
Knight vs Snail!
A snail is invited to dine with a reed situated one league away. However, in one day it can travel only one ounce of a foot. Tell me, whoever wishes, how long it will take the snail to reach its meal.
Solution: In one league there are 1,500 paces, or 7,500 feet, or 90,000 ounces of a foot. Since the snail travels one ounce per day, the time required is 246 years and 210 days.
Proposition 2 – Of a Walker
In the street, a walker meets some men and says to them: “I wish there were as many others as there are of you, plus half of half, then half of that last number. Then, with me included, we would be one hundred.” Tell me, whoever wishes, how many men the walker met.
Solution: The walker met 36 men. Adding 36 gives 72. Half of half of 72 is 18. Half of 18 is nine. Thus 72 and 18 make 90; adding nine gives 99. With the walker himself, that makes 100.
Proposition 3 – Of Two Walkers
Two men were walking. Seeing some storks, they wondered how many there were. After discussion they said: “If there were three times this number, and if half of a third of that number were added, and then two more came, there would be one hundred.” Who can tell how many storks there are?
Solution: There are 28 storks. Three times 28 makes 84; half of a third of 84 is 14. That gives 98 storks, and if two more arrive, there are 100.
Proposition 4 – Of a Man and Some Horses
A man saw some horses grazing in a meadow and said: “Heaven grant that these horses were mine! If I had as many as there are, plus half of half, I should have one hundred.” Tell me, whoever wishes, how many horses are in the meadow.
Solution: There are 80 horses grazing. If the man had as many, he would have 80. Half of half of 80 is 20; adding 20 to 80 makes 100.
Proposition 5 – Of a Buyer of Pigs
12th century manuscript from England showing two pigs and a man with an axe – British Library MS Lansdowne 383 f. 8
A merchant said: “I wish to buy one hundred pigs for 100 deniers. A boar costs 10 deniers, a sow five, and two piglets one denier.” Who can determine how many boars, sows, and piglets the merchant will buy with his hundred deniers?
Solution: The merchant can buy one boar and nine sows for 55 deniers. He then buys ninety piglets, priced at 5 deniers for ten, which makes 45 deniers. Thus he has one hundred pigs for 100 deniers.
Proposition 6 – Of Two Merchants
Two merchants shared 100 ducats to buy pigs. At the rate of two ducats for five pigs, they bought their herd intending to fatten and resell it for profit. Unfortunately, it was not the right season to fatten pigs, and they did not wish to feed them through the winter. They tried to sell at a profit but could not. As they could not sell at a higher price, they agreed to divide the pigs and sell them themselves — yet they managed to make a profit. Who can tell how they divided them, for how much they sold them, and what profit they made?
Solution: With their 100 ducats they bought 250 pigs (2 ducats for five). Each took 125 pigs: one chose the smaller, the other the larger. The smaller ones were sold in groups of three for one ducat, the larger in groups of two for one ducat — still two ducats for five pigs.
The first sold 120 pigs for 40 ducats; the second 120 for 60 ducats. Each had five pigs left, which they sold for four ducats and two deniers — their profit.
Note. The sale of 120 pigs each brought 100 ducats in total. One sold his remaining five pigs at one ducat for three pigs (1 ⅔ ducats); the other sold his at one ducat for two pigs (2 ½ ducats). Together this makes 4 ⅙ ducats. Since one ducat equals 12 deniers, the profit is 4 ducats and 2 deniers.
Proposition 7 – Of a Disk
There is a disk weighing 30 pounds and costing 600 ducats. It is composed of gold, silver, brass, and tin. There is three times as much silver as gold; three times as much brass as silver; and three times as much tin as brass. Who can determine the weight of each metal? Note. One pound equals 12 ounces.
Solution: Gold weighs 9 ounces; silver three times as much, i.e. 2 pounds 3 ounces; brass three times as much as silver, i.e. 6 pounds 9 ounces; and tin three times as much as brass, i.e. 20 pounds 3 ounces — making 30 pounds in all.
Alternative solution: Gold costs 15 ducats; silver three times as much (45); brass three times 45 (135); tin three times 135 (405). The total is 100 ducats, corresponding to 30 pounds.
Proposition 8 – Of a Barrel
Initial ‘V’: a monk-cellarer tasting wine from a barrel whilst filling a jug. Sloan 2435 f.44v
There is a barrel holding 100 metretai, connected to three pipes from as many openings. The first pipe fills a third and a sixth of the barrel; the second a third; the third a sixth. Tell me, whoever wishes, how many setiers flow through each pipe. Note. One metrētēs equals 72 setiers.
Solution: The first pipe lets through 3,600 setiers of liquid, the second 2,400, and the third 1,200.
Proposition 9 – Of a Blanket
I have a blanket 100 cubits long and 80 wide. I wish to cut it into pieces to make cloaks. Each cloak requires a piece 5 cubits long and 4 wide. Tell me, I beg you, wise man, how many cloaks can be made from this blanket. Note. One cubit equals 44.46 cm, or 1½ feet.
Solution: The 80th part of 400 is 5; the 100th part is 4. That is, 80 times 5 or 4 times 100 — thus 400 cloaks can be made.
Proposition 10 – Of a Linen Cloth
“I have a piece of cloth 60 cubits long and 40 wide. I wish to cut it into pieces 6 cubits long and 4 wide to make tunics.” Tell me, whoever wishes, how many tunics can be made.
Solution: The tenth part of 60 is 6; the tenth part of 40 is 4. Thus one-tenth of 60 or one-tenth of 40; multiplying by 10 gives 100 pieces of 6 by 4 cubits — hence 100 tunics.
Dr Lorris Chevalier, who has a Ph.D. in medieval literature, is a historical advisor for movies, including The Last Duel and Napoleon. Click here to view his website.
Try solving ten medieval math riddles first posed by the scholar Alcuin of York. These puzzles from the early Middle Ages reveal how logic and learning were taught in Charlemagne’s time — and they’re still tricky to solve today.
By Lorris Chevalier
Albinus Flaccus Alcuin (735–804), a monk, scholar, and teacher, was among the most learned minds of the early Middle Ages. Appointed by King Charlemagne (c. 742–814) as tutor and adviser, he played a key role in reforming the educational system of the Carolingian Empire. In addition to his theological and pedagogical writings, Alcuin compiled a fascinating Latin collection entitled Propositiones ad acuendos juvenes — Problems to Sharpen the Wits of the Young.
This remarkable work presents 53 recreational puzzles: 33 on arithmetic (including eight dealing with division of goods), 12 on geometry, and eight on logic (four concerning river crossings). As noted by Pierre Dedron and Jean Itard in Mathématiques et mathématiciens, it represents “one of the earliest examples of mathematical recreations.”
Each problem is accompanied by its solution. Typically, Alcuin begins by stating the answer and then provides a brief justification. Of course, the mathematical tools and methods of his time did not allow for the kind of detailed reasoning we expect in modern solutions — yet his ingenuity remains striking even today.
Proposition 1 – Of a Snail
A snail is invited to dine with a reed situated one league away. However, in one day it can travel only one ounce of a foot. Tell me, whoever wishes, how long it will take the snail to reach its meal.
Solution: In one league there are 1,500 paces, or 7,500 feet, or 90,000 ounces of a foot. Since the snail travels one ounce per day, the time required is 246 years and 210 days.
Proposition 2 – Of a Walker
In the street, a walker meets some men and says to them: “I wish there were as many others as there are of you, plus half of half, then half of that last number. Then, with me included, we would be one hundred.” Tell me, whoever wishes, how many men the walker met.
Solution: The walker met 36 men. Adding 36 gives 72. Half of half of 72 is 18. Half of 18 is nine. Thus 72 and 18 make 90; adding nine gives 99. With the walker himself, that makes 100.
Proposition 3 – Of Two Walkers
Two men were walking. Seeing some storks, they wondered how many there were. After discussion they said: “If there were three times this number, and if half of a third of that number were added, and then two more came, there would be one hundred.” Who can tell how many storks there are?
Solution: There are 28 storks. Three times 28 makes 84; half of a third of 84 is 14. That gives 98 storks, and if two more arrive, there are 100.
Proposition 4 – Of a Man and Some Horses
A man saw some horses grazing in a meadow and said: “Heaven grant that these horses were mine! If I had as many as there are, plus half of half, I should have one hundred.” Tell me, whoever wishes, how many horses are in the meadow.
Solution: There are 80 horses grazing. If the man had as many, he would have 80. Half of half of 80 is 20; adding 20 to 80 makes 100.
Proposition 5 – Of a Buyer of Pigs
A merchant said: “I wish to buy one hundred pigs for 100 deniers. A boar costs 10 deniers, a sow five, and two piglets one denier.” Who can determine how many boars, sows, and piglets the merchant will buy with his hundred deniers?
Solution: The merchant can buy one boar and nine sows for 55 deniers. He then buys ninety piglets, priced at 5 deniers for ten, which makes 45 deniers. Thus he has one hundred pigs for 100 deniers.
Proposition 6 – Of Two Merchants
Two merchants shared 100 ducats to buy pigs. At the rate of two ducats for five pigs, they bought their herd intending to fatten and resell it for profit. Unfortunately, it was not the right season to fatten pigs, and they did not wish to feed them through the winter. They tried to sell at a profit but could not. As they could not sell at a higher price, they agreed to divide the pigs and sell them themselves — yet they managed to make a profit. Who can tell how they divided them, for how much they sold them, and what profit they made?
Solution: With their 100 ducats they bought 250 pigs (2 ducats for five). Each took 125 pigs: one chose the smaller, the other the larger. The smaller ones were sold in groups of three for one ducat, the larger in groups of two for one ducat — still two ducats for five pigs.
The first sold 120 pigs for 40 ducats; the second 120 for 60 ducats. Each had five pigs left, which they sold for four ducats and two deniers — their profit.
Note. The sale of 120 pigs each brought 100 ducats in total. One sold his remaining five pigs at one ducat for three pigs (1 ⅔ ducats); the other sold his at one ducat for two pigs (2 ½ ducats). Together this makes 4 ⅙ ducats. Since one ducat equals 12 deniers, the profit is 4 ducats and 2 deniers.
Proposition 7 – Of a Disk
There is a disk weighing 30 pounds and costing 600 ducats. It is composed of gold, silver, brass, and tin. There is three times as much silver as gold; three times as much brass as silver; and three times as much tin as brass. Who can determine the weight of each metal? Note. One pound equals 12 ounces.
Solution: Gold weighs 9 ounces; silver three times as much, i.e. 2 pounds 3 ounces; brass three times as much as silver, i.e. 6 pounds 9 ounces; and tin three times as much as brass, i.e. 20 pounds 3 ounces — making 30 pounds in all.
Alternative solution: Gold costs 15 ducats; silver three times as much (45); brass three times 45 (135); tin three times 135 (405). The total is 100 ducats, corresponding to 30 pounds.
Proposition 8 – Of a Barrel
There is a barrel holding 100 metretai, connected to three pipes from as many openings. The first pipe fills a third and a sixth of the barrel; the second a third; the third a sixth. Tell me, whoever wishes, how many setiers flow through each pipe. Note. One metrētēs equals 72 setiers.
Solution: The first pipe lets through 3,600 setiers of liquid, the second 2,400, and the third 1,200.
Proposition 9 – Of a Blanket
I have a blanket 100 cubits long and 80 wide. I wish to cut it into pieces to make cloaks. Each cloak requires a piece 5 cubits long and 4 wide. Tell me, I beg you, wise man, how many cloaks can be made from this blanket. Note. One cubit equals 44.46 cm, or 1½ feet.
Solution: The 80th part of 400 is 5; the 100th part is 4. That is, 80 times 5 or 4 times 100 — thus 400 cloaks can be made.
Proposition 10 – Of a Linen Cloth
“I have a piece of cloth 60 cubits long and 40 wide. I wish to cut it into pieces 6 cubits long and 4 wide to make tunics.” Tell me, whoever wishes, how many tunics can be made.
Solution: The tenth part of 60 is 6; the tenth part of 40 is 4. Thus one-tenth of 60 or one-tenth of 40; multiplying by 10 gives 100 pieces of 6 by 4 cubits — hence 100 tunics.
You can read more of these puzzles in the new book Alcuin’s Recreational Mathematics: River Crossings and other Timeless Puzzles, by Marcel Danesi
Dr Lorris Chevalier, who has a Ph.D. in medieval literature, is a historical advisor for movies, including The Last Duel and Napoleon. Click here to view his website.
Click here to read more from Lorris Chevalier
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