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> apart from ingenuity, clever guesswork That's what I'm asking though: what is the ingenuity or cleverness required to solve this problem? The only technique

> apart from ingenuity, clever guesswork That's what I'm asking though: what is the ingenuity or cleverness required to solve this problem? The only technique I can think of is guess and check.

Anderkent 12y ago
In this particular case it's pretty easy because you know you'll be working with small numbers, so once you establish tighter bounds the work to 'guess and check' will be small. This 'bound tightening' is the 'ingenuity' required for the problem. There's no general process to achieve that, which is why it requires ingenuity :) SPOILERS BELOW To do that you have to notice the common factor in the initial equations. Let n1, n2, and n3 be number of chickens sold pre-lunch time by each farmer. The initial equations are: n1*x + (10-n1)*y = 35 n2*x + (16-n2)*y = 35 n3*x + (26-n3)*y = 35 Transforming them a little: n1 * (x - y) + 10y = 35 ;[1] n2 * (x - y) + 16y = 35 ;[2] n3 * (x - y) + 26y = 35 ;[3] The common factor of (x-y) looks useful; you can establish relationships between n1, n2 and n3 with it: (n1 - n2) * (x - y) - 6y = 0 ;([1] - [2]) (n1 - n3) * (x - y) - 16y = 0 ;([1] - [3]) (n1 - n2) = 6 * (y / (x - y)) ;[4] (n1 - n3) = 16 * (y / (x - y)) ;[5] (n1 - n2) = 3/8 (n1 - n3) ;[6] If you get there the puzzle is basically solved but for a bit of number crunching - you know n1, n2 and n3 are positive integers, n1 <= 10, so for [6] either n1=n2=n3 (trivial 0-post-lunch-price solution) or n1 - n2 = 3, n1 - n3 = 8. From there you substitute back into [4] and get 3 * (x-y) = 6y, so x = 3y. You now have simple relations between n1, n2 and n3; as well as between x and y. The remaining step is to tie x or y to one of the n's by substituting into initial equations. For example from [1]: n1 * 2y + 10y = 35 2y = 35 / (n1 + 5) y must be integral in pennies; that part is tricker to derive and the simplest solution is to just notice that since n1 >= 0, and n3 >= 0, and n1 = n3 + 8, then 8 <= n1 <= 10, and try 8, 9 and 10 for n1. With n1 = 9, you get y = 1.25. That's the only 'guess and check' part of the problem.
CarolineW 12y ago
Quite often swathes of the search space can be eliminated or winnowed by using modulo arguments. You can also sometimes fold these problems into smaller spaces by considering equivalence arguments. There are many ad hoc techniques that no one has yet classified or unified, they are scattered across many mathematical disciplines.

> apart from ingenuity, clever guesswork That's what I'm asking though: what is the ingenuity or cleverness required to solve this problem? The only technique I can think of is guess and check.

Anderkent 12y ago

In this particular case it's pretty easy because you know you'll be working with small numbers, so once you establish tighter bounds the work to 'guess and check' will be small. This 'bound tightening' is the 'ingenuity' required for the problem. There's no general process to achieve that, which is why it requires ingenuity :) SPOILERS BELOW To do that you have to notice the common factor in the initial equations. Let n1, n2, and n3 be number of chickens sold pre-lunch time by each farmer. The initial equations are: n1*x + (10-n1)*y = 35 n2*x + (16-n2)*y = 35 n3*x + (26-n3)*y = 35 Transforming them a little: n1 * (x - y) + 10y = 35 ;[1] n2 * (x - y) + 16y = 35 ;[2] n3 * (x - y) + 26y = 35 ;[3] The common factor of (x-y) looks useful; you can establish relationships between n1, n2 and n3 with it: (n1 - n2) * (x - y) - 6y = 0 ;([1] - [2]) (n1 - n3) * (x - y) - 16y = 0 ;([1] - [3]) (n1 - n2) = 6 * (y / (x - y)) ;[4] (n1 - n3) = 16 * (y / (x - y)) ;[5] (n1 - n2) = 3/8 (n1 - n3) ;[6] If you get there the puzzle is basically solved but for a bit of number crunching - you know n1, n2 and n3 are positive integers, n1 <= 10, so for [6] either n1=n2=n3 (trivial 0-post-lunch-price solution) or n1 - n2 = 3, n1 - n3 = 8. From there you substitute back into [4] and get 3 * (x-y) = 6y, so x = 3y. You now have simple relations between n1, n2 and n3; as well as between x and y. The remaining step is to tie x or y to one of the n's by substituting into initial equations. For example from [1]: n1 * 2y + 10y = 35 2y = 35 / (n1 + 5) y must be integral in pennies; that part is tricker to derive and the simplest solution is to just notice that since n1 >= 0, and n3 >= 0, and n1 = n3 + 8, then 8 <= n1 <= 10, and try 8, 9 and 10 for n1. With n1 = 9, you get y = 1.25. That's the only 'guess and check' part of the problem.

CarolineW 12y ago

Quite often swathes of the search space can be eliminated or winnowed by using modulo arguments. You can also sometimes fold these problems into smaller spaces by considering equivalence arguments. There are many ad hoc techniques that no one has yet classified or unified, they are scattered across many mathematical disciplines.