# A cute problem that illustrates the power of integers

The following problem was introduced to me as a ‘homework problem’ for the child of a certain genius-level mathematician, which is likely apocryphal. Nevertheless, the problem itself is interesting.

Problem: Three farmers brought 10, 16, and 26 chickens to the market to sell. To avoid a price war, they agreed to sell each chicken for the same price. By midday, none of them had sold all of their chickens. They then agreed to lower their prices by the same amount in order to unload all of their chickens. By the end of the day, each farmer sold all of the chickens they brought to the market and each made 35 dollars. What is the difference of the morning price and the afternoon price?

This question seems like it does not provide enough information. Indeed it does not, if one fails to consider that certain quantities (namely the amount of chickens) are integers. To wit, let us establish the solution as follows.

Let $x,y,z$ respectively denote the number of chickens sold by farmer 1, 2, and 3 in the morning. Note that $x,y,z$ are necessarily integers. Let $C$ denote the price of a chicken in the morning, and $C'$ denote the price of a chicken in the afternoon. Let $D = C'/(C - C')$. We then obtain the following equations:

$Cx + (10 - x)C' = 35,$

$Cy + (16 - y)C' = 35,$ and

$Cz + (26-z)C' = 35.$

Subtracting the third equation from the first and second and re-arranging, we obtain

$x - z = 16D, y - z = 10D.$

Now the fact that $x,y,z$ are positive integers really come in handy. In particular, we must have $16D, 10D$ are both integers. The first condition implies that the denominator of $D$ must be a power of 2, and the second implies that it must in fact be exactly 2. Now we write $D = G/2$, where $G$ is a positive odd integer. Now we examine the equation

$x = 8G + z.$

Note that since Farmer 1 did not sell all of his chickens in the morning, we must have

$1 \leq x \leq 9$.

However, $G \geq 1$ and $z \geq 1$, whence $8G + z \geq 9$, and so $x = 9$ is the only possibility. Thus, $z = 1$ and $G = 1$. This shows that

$y = 5G + 1 = 6.$

With this information in hand, we go back to examine the equation

$9C + C' = 35, 6C + 10C' = 35$.

Substituting, we see that $C = 15/4, C' = 5/4$. So it would appear that $C - C' = 15/4 - 5/4 = 5/2$.

But wait! We had only shown that the values of $C, C'$ above satisfy two of the three required equations, so for due diligence we must examine the final equation

$C + 25C' = 35$.

But

$15/4 + 25(5/4) = 140/4 = 35,$

so indeed $C = 15/4, C' = 5/4$ is an acceptable solution.

Note that the solution really required the fact that $x,y,z$ were positive integers and constrained in a specific way. Without these (seemingly unimportant) tidbits of information, the problem would be intractable.