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The following surprisingly difficult challenge problem was given to the students at my son's Math Circle last week. Unlike the three princesses puzzle in my previous post, this one does require a bit of pen and paper to solve; I spent more than fifteen minutes on the problem, during half of which I was convinced the problem was ill-posed. Remarkably, one of the students in the class actually solved it in class time (not my son, though).

As before, I would prefer if you not simply spoil the answer in comments, but instead discuss your thought processes in how you arrived at your solution.

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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?

As before, I would prefer if you not simply spoil the answer in comments, but instead discuss your thought processes in how you arrived at your solution.

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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?

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- As some people have shown in the comments there are 3 possible solutions to the problem (3 tuples of before/after prices). There is only one solution if you assume that the solution must be integer in cents. But the problem does not state this. If you found only one solution, it might be fun to go back and try to find out which assumption made you miss the other 2.Dec 23, 2014
- it took me 30 minutes, so my IQ is only half of Terence's at best.Dec 25, 2014
- My common lisp solution https://gist.github.com/arademaker/1717903c5ae196586721Feb 5, 2015
- Wow, finally solved it. I was desperately looking for a method and eventually decided to set up simultaneous equations. I went through multiple simultaneous equations and just as I was going to give up I stumbled across this simultaneous equation that I write down being: 9x + y = 35 and 6× + 10y = 35

Here I assumed the first farmer sold his 10 chicken 9 before lunchtime and 1 after as represented in the first equation. I also assumed the second farmer sold his 16 chickens, 6 before lunchtime and 10 after as represented in the second equation. X represents the first unknown price and Y represents the second unknown price. The solution to this simultaneous equation is x= 3.75 and y= 1.25. Now I looked at the third farmer and assumed the first possibility of him selling his 26 chicken 1 before lunchtime and 25 after. Finally I substituted the x and y values into this assumption so it looks like this 1 (3.75)+ 25 (1.25)=35. This confirmed that before lunch the farmers 1-3 sold their chickens 9, 6 ,1 and after lunch 1,10,25 and they all sold their chickens at firstly $3.75 then $1.25.Feb 8, 2015 - Find an integer between 0 and 10 that, if multiplied by 5/3, is still and integer between 0 and 10. 3 seems to be our only choice. Thus, before lunch, the first farmer sells 3 more than the second farmer, who sells 5 more than the third farmer. We also know from this that the price after lunch is one-third of the price before lunch. In all, (10,7,2) (9,6,1) and (8,5,0) are feasible solutions of sales before lunch if we don't consider the restriction of prices(accurate to cents). Trying out these three, (9,6,1)with prices 3.75 and 1.25 is what we need. Not much calculation, actually.Feb 26, 2015
- $35 in the morning and each farmer sold one chicken. In the afternoon they give away the remaining stock.

Hey, if central banks can set interest rates at less than zero, farmers can "sell" chickens for $0. :)Apr 23, 2015