# An arithmetic puzzle from 1645

The following problem is found in the first Norwegian arithmetic, published in 1645 by Tyge Hansøn:

Three hundred oxen large and small,
A cattle owner wanted to buy:
3 for 63 daler he got.
Again, he let them sell,
3 for 63 daler they fetched,
both slim and fat.
787$1\over 2$ daler was his profit.
Tell me, how did that happen?
Whoever wants to calculate that, consider rightly
and argue the issue well.
Then it becomes quite simple.
to calculate for a farmer.

So apparently, the cattle owner bought and sold for the same price, but he still made a profit. There is no solution to this problem in the text. I have a puzzle-like solution, but would like to hear other suggestions before I reveal it.

Some notes:
The original is written in verse, with rhymes. I have translated the text into modern English, but had to sacrifice some of the poetic effects in order to preserve the factual content.

In the coinage system used in Denmark-Norway at the time, one daler could be divided into 96 skillings. The problem is found in a chapter on the rule of three, so a solution should in some way be true to the principle of proportionality. Hence, any suggestion that the herd has increased in size, or similar ideas, would not be satisfactory.

And if anyone has seen a similar problem, I would be very interested in hearing about it.

• What if you split Large vs. Small from Fat vs. Slim such that there were Large Slim oxen or Small Fat ones? Could he just have redefined the groups? – Bobson Jun 30 '14 at 18:46
• Also, see this book and see if it has an answer? I can't read it, but I can tell that it's referring to the same problem. – Bobson Jun 30 '14 at 18:48
• @JoeZ. Of course you have to do something fishy in order to get a profit out of nothing. :) – Per Manne Jun 30 '14 at 18:52
• @Bobson Tyge Hansøn's book is titled Arithmetica Danica. Only one copy is known to exist today; it can be found in the national library of Denmark. And the teacher's magazine is called Tangenten; it's a Norwegian language journal focused on the teaching of mathematics. – Per Manne Jun 30 '14 at 21:17
• It hasn't been mentioned and it seems like cheating but could the solution be 1/8th of them gave birth? – kaine Jul 1 '14 at 12:51

The problem in modern English might be stated as follows:

A farmer wanted to buy three hundred oxen. He bought them at a price of $3$ oxen for $\$63$. Afterwards, he sold the oxen at$3$for$\$63$ as well, but managed to make a profit of $\$787.50$. How did he manage this? At first blush, as you said. it appears that the farmer should have left with exactly as much money as he'd come with - after all, he bought and sold the oxen at$\$21$ an ox.

First of all, let's find the price that $300$ oxen would have fetched. Since you mentioned the rule of three, we'll use that to calculate it. Let $P$ be the total amount of money the farmer spend to buy the cows:

$$\frac {\63}{3} = \frac{P}{300}$$

$$P = \frac{63 \times 300}{3} = \6300$$

Note that $\$787.50$is exactly one-eighth of$\$6300$. Somehow, when selling the oxen, he received 9 dollars for every 8 dollars he spent, even selling them at the same rate.

I can only imagine that the answer has something to do with currency exchange and the deceptive use of the word "daler" in the original riddle, and that another Nordic country that used dalers at the time divided them into $108$ skillings instead of $96$.

If this were the case and the farmer had done his entire bovine-selling operation in skillings, he could have bought the oxen for $21 \times 96$ skillings apiece in one country and sold them for $21 \times 108$ skillings in another country, making a profit of $21 \times 12$ skillings per ox for a total of $\frac{21 \times 12 \times 300}{96} = 787.5$ dalers in profit.

Doing some cursory research on Wikipedia reveals that Norway/Denmark had two dalers - one worth 96 skillings and one worth 120. This still works as a solution as above – if the farmer sold half the cows for those 120-skilling dalers, he would have made the same profit of $\frac{21 \times 24 \times 150}{96} = 787.5$ dalers.

• An interesting solution (+1)! Unfortunately, the Wikipedia reference is not correct - the daler with 120 skillings did not exist prior to 1816. The second daler was the courant daler, which was introduced in 1695 as a calculation device to reflect that the skilling coins were debased, and did not contain as much silver as their face value would indicate. It was never minted, and did not exist in 1645 in any case. – Per Manne Jun 30 '14 at 13:35
• Unfortunately that was the only thing that came to mind. – Joe Z. Jun 30 '14 at 17:48
• Also, would you technically have a "profit" just because you have a higher number of a weaker currency? After all it's worth the same amount in the exact same currency – musefan Jul 1 '14 at 14:36
• No; the point is that you bought the cows in the weaker currency and sold them in the stronger one, thus making a profit. – Joe Z. Jul 1 '14 at 17:58

Following up on the suggestion by Dennis Meng, I'm posting my suggestion as an answer instead of as part of the question.

So here are my thoughts on the problem. (1) As Joe Z comments, the cattle owner needs to increase prices by one eighth in order to make the stated profit. (2) For some reason, the price is given as 63 daler for 3 oxen, rather than the simpler 21 daler for one ox. (3) There seems to be two different kinds of oxen, where some are big and fat, and some are small and slim.

Suggestion: Let there be different prices for large and small oxen. The farmer buys 200 large oxen and 100 small oxen, where two large oxen and one small cost 63 daler together. He increases both prices by one eighth and sells the herd. The new prices happen to be such that one large and two small oxen cost 63 daler together!

Working out the details, large oxen were bought for $23 {1\over 3}$ daler each and sold for $26{1\over 4}$, whereas small oxen were bought for $16{1\over 3}$ daler each and sold for $18{3\over 8}$. Note that the fractions cause no problems, as there are 96 skilling in a daler.

I don't know if this is the intended solution, but it seems consistent with both the statement of the problem and the principle of proportionality, which is present in all rule of three computations. I guess the only way to find out for sure is if the problem was taken from another source, with a fuller discussion.

• The problem with this answer is that it requires you to buy 100 small oxen and then sell 200. – namey Apr 6 '15 at 22:23
• I think Bobson's comment for the question leads to the correct answer: The farmer bought 200 large oxen and 100 small oxen. He then sold 100 fat oxen and 200 slim oxen. The rest of the details are as in this answer. This is in the spirit of the question and makes sense if there were 100 large fat oxen, 100 large slim oxen and 100 small slim oxen. – namey Apr 6 '15 at 22:27

It appears that we are missing a label here. "3 for 63" does not necessarily denote the number of oxen. Perhaps it refers to the weight of the oxen. The farmer could have bought the oxen at a set price, fattened them up, then sold the same number of oxen "at the same price" for a profit.