4 ms·

Since this puzzle has a definitive answer, it is solvable by simply iterating from 1 to 10 for the number of chickens sold by the 1st farmer and then seeing whi

by eps 12y ago

Since this puzzle has a definitive answer, it is solvable by simply iterating from 1 to 10 for the number of chickens sold by the 1st farmer and then seeing which quantity can yield integer quantities for other farmers. And since it's for kids and the number is 10 and I would guess it's the way this puzzle was meant to be solved (in the Math Circle). How to solve it or to even determine it's uniquely solvable for arbitrary set of 10, 16, 26 and 35 - I don't know, but I'm pretty sure it's doable and has to do with the groups :)

lazyjones 12y ago
There doesn't seem to be an integer solution though. I did the following: - starting from farmer with 10 chicken, I checked for i = 0..5 (number of chicken possibly sold in the worse part of the day) whether a = 4,5,6,7,8... were working prices for the first half of the day, by calculating 35 - a * (10 - i) and checking whether the result was divisible by an integer in the range 1 .. a-1 (can be done quickly by writing down a table) - with possible solutions for the first merchant (e.g. a = 7, b = 2, sold 3 for a and 7 for b), repeat for 2nd and 3rd merchant and find solutions that work there also (can limit the choices of a to prior solutions from merchant 1) - (also, the second price must be 1 if integer due to 26 chicken sold at $35) I'd assume that I either did some calculations wrong (my old head) or the teacher would at least either expect the kids to also check non-integer solutions (perhaps try 1.5, 2.5 etc.) or actually solve this: a * x1 + b * (10-x1) = 35 a * x2 + b * (16-x2) = 35 a * x3 + b * (26-x3) = 35 a > b x1 < 5 (unsure) x2 < 8 (unsure) x3 < 13 (unsure) (x1, x2, x3 positive integers; a, b at least rational numbers?)
- patio11 12y ago
  You made an assumption that I originally did, which is that farmer $FOO was dissatisfied with his sales because he had half of his stock unsold at the midday mark. This assumption is unsupported by the text of the problem. Try relaxing it. Also, chickens are integers but prices are not is a core insight. Life gets easier if you solve a related problem which forces prices to be integers as well, such as an alternate universe where chickens are priced in nickels.
- soegaard 12y ago
  Use cents as units to get an integer problem.

by eps 12y ago

lazyjones 12y ago

There doesn't seem to be an integer solution though. I did the following: - starting from farmer with 10 chicken, I checked for i = 0..5 (number of chicken possibly sold in the worse part of the day) whether a = 4,5,6,7,8... were working prices for the first half of the day, by calculating 35 - a * (10 - i) and checking whether the result was divisible by an integer in the range 1 .. a-1 (can be done quickly by writing down a table) - with possible solutions for the first merchant (e.g. a = 7, b = 2, sold 3 for a and 7 for b), repeat for 2nd and 3rd merchant and find solutions that work there also (can limit the choices of a to prior solutions from merchant 1) - (also, the second price must be 1 if integer due to 26 chicken sold at $35) I'd assume that I either did some calculations wrong (my old head) or the teacher would at least either expect the kids to also check non-integer solutions (perhaps try 1.5, 2.5 etc.) or actually solve this: a * x1 + b * (10-x1) = 35 a * x2 + b * (16-x2) = 35 a * x3 + b * (26-x3) = 35 a > b x1 < 5 (unsure) x2 < 8 (unsure) x3 < 13 (unsure) (x1, x2, x3 positive integers; a, b at least rational numbers?)

patio11 12y ago
You made an assumption that I originally did, which is that farmer $FOO was dissatisfied with his sales because he had half of his stock unsold at the midday mark. This assumption is unsupported by the text of the problem. Try relaxing it. Also, chickens are integers but prices are not is a core insight. Life gets easier if you solve a related problem which forces prices to be integers as well, such as an alternate universe where chickens are priced in nickels.
soegaard 12y ago
Use cents as units to get an integer problem.