#COW AND CHICKEN MATH PROBLEM TRIAL#

Similar such experimentation will uncover one solution where the farmer buys 4 cows, one where he buys 3 cows, one where he buys 2 cows, and one where he buys 1 cow. To recap, if the farmer buys 4 cows, 1 sheep, 3 chickens, and 92 rabbits, it satisfies all of the requirements. Perfect! If he exchanges two rabbits for sheep, that means an increase of $5 which is exactly what we needed.

At this point, I realized that if he bought a chicken instead of a sheep, he'd 'earn back' $0.50 but spend $3, for a net increase of $2.50 to his expenditures. Buying 94 rabbits would give the desired total of 100 animals, but also put him 5 dollars over budget. I figured the best way to go from here was to buy as many rabbits as possible because they're cheapest. Then I looked at what would happen if he bought 4 cows and 1 of each other animal. I began by using the method I outlined in my original post and worked down until I realized that if the farmer bought 5 cows, he could not buy the required other 95 animals while sticking to the budget and the requirement that he buy at least one of each animal.

#COW AND CHICKEN MATH PROBLEM TRIAL#

I found another solution very quickly using a bit of trial and error. You identified one such solution, which should suffice (although I have seen teachers refuse to accept any answer except their 'correct' answer, even if the other solution(s) fully satisfy the problem's criteria. Which method do you prefer? Which one would you use with your students? Share yours with the rest of us.As it turns out, there's four unique solutions to this problem. A consequence of the theorem is that there are exactly positive integers which cannot be expressed in the form. With Chris's algebraic solution, we're sharing four methods of solution to the chickens-and-cows problem. The Chicken McNugget Theorem (or Postage Stamp Problem or Frobenius Coin Problem) states that for any two relatively prime positive integers, the greatest integer that cannot be written in the form for nonnegative integers is. Chickens Cows Number of legs ( ) × 2 × 4 2 184 2. How many chickens and cows have 8 legs and heads altogether?Ĭlearly, 1 chicken and 1 cow have 8 legs and heads altogether. Grasp the relation Let represent the number of cows, and be a variable quantity that is much less than. Then each group of 2 chickens and 1 cow would have a total of = 11 legs and heads.ġ6 groups of 2 chickens and 1 cow would have a total of 16 × 11 = 176 legs and heads. Let the number of cows be x.Then, number of hens would be (180 - x) According to the question(180 - x) x 2 + 4x 420 360 - 2x + 4x 420 2x 420 - 360. Suppose there were exactly twice as many chickens as cows. Method 3 ( Using the 'Make a supposition' strategy) The Chicken McNugget Theorem (or Postage Stamp Problem or Frobenius Coin Problem) states that for any two relatively prime positive integers, the greatest integer that cannot be written in the form for nonnegative integers is.

Unless there's a more elegant Sakamoto solution, I find that it doesn't differ much conceptually from its algebraic cousin.

Cow and chicken math problem