# Terry Tao shares a math circle problem

I saw Patrick Honner tweet about Terry Tao’s math circle post last night:

Hopefully the link in the tweet to the original post works, but in case it doesn’t Tao’s post on google+ is here:

https://plus.google.com/u/0/114134834346472219368/posts/CR1ZoNe9ojQ%20

and here’s the problem itself:
“Three farmers were selling chickens at the local market.  One farmer had 10 chickens to sell, another had 16 chickens to sell, and the last had 26 chickens to sell.  In order not to compete with each other, they agreed to all sell their chickens at the same price.  But by lunchtime, they decided that sales were not going so well, and they all decided to lower their prices to the same lower price point.  By the end of the day, they had sold all their chickens.  It turned out that they all collected the same amount of money, \$35, from the day’s chicken sales.  What was the price of the chickens before lunchtime and after lunchtime?”

In his post Tao asks not for the solution but for the thought process you went through to solve the problem.  I like these “thought process” posts, so I thought I’d give another one a shot.  My first one was here:

https://mikesmathpage.wordpress.com/2014/08/12/problem-solving/%20

For this math circle problem I made the same assumptions that many of the commentators on Tao’s blog were making -> (i) the number of chickens sold by each of the farmers in both the morning and the afternoon was not zero, and (ii) that the both the morning and afternoon prices were an integer number of cents.

The first thing I did was look for a simple example just to get my arms around the problem.  With a morning price of \$4 per chicken and an afternoon price of \$1 per chicken you can get close to the situation the problem describes.  I didn’t put much thought into those two prices – I just chose those two prices because \$4 was above the average price for the 10 chickens and \$1 was below the average price of the 26 chickens.

Farmer 1:  8 chickens at \$4 plus 2 chickens at \$1 = \$34
Farmer 2:  6 chickens at \$4 plus 10 chickens at \$1 = \$34

Farmer 3:  3 chickens at \$4 plus 23 chickens at \$1 = \$35

This short exercise gave me some hope that this initial guess was pretty close to the answer to the problem.

Next I wrote down some equations.   Call the morning price x and the afternoon price y, and let A, B, and C represent the number of chickens sold at the morning price by Farmers 1,2, and 3 respectively.  We have (in cents):

(1) Ax + (10 – A)y = 3500

(2) Bx + (16 – B)y = 3500

(3) Cx + (26 – C)y = 3500

A little equation combining, namely (1) + (2) – (3), yields the equation:

(4) (A + B – C)*(x – y) = 3500

Since both terms on the left hand side are integers the problem now is to find the right factors of 3500.  Fortunately 3500 doesn’t have too many factors, and double fortunately the example from the beginning tells me to expect (A + B – C) to have a value around 11.

There are two factors of 3500 that are on either side of 11, namely 10 and 14.  The values of (x – y) for those two choices are 350 and 250.  I checked 350 first.

Rewriting equation (3) you can get:

(5) C*(x – y) + 26y = 3500.

When (x – y) is 350 this equation becomes 350C + 26y = 3500, or

(6)  26y = 350*(10 – C).

This equation will not have any solutions when both y and C are positive integers since the left hand side is divisible by 13 and the right hand side isn’t.

Bad luck, I suppose, but we do have the second potential solution from above when (A + B – C) = 14 and the price change from morning to afternoon is 250 cents.  As Tao has asked to not give away the solution to the problem, I won’t work through that math but will say you can find a way to make both sides of the equation similar to (6) be divisible by 13 here.  Yay!

Definitely a fun challenge problem.  Unlike the Tim Gowers’s IMO problem that inspired my first “thought process” post, I hope to use this problem for a neat little Family Math project with my kids later this week!