Bernie has three pastures, each containing an equal number of animals. He sold all the animals of his pastures to 37 persons, each buying an equal number of animals.
If the cows cost $32.00 each, goats $15.00 each and pigs $5.00 each - determine the maximum number of animals Bernie could have, if he received $1661.00 from the sale.
(In reply to
Solution by Larry)
Larry:
Your first analysis is elegant, proving that the solution (if one exists) must be either 111 or 222.
It is not hard to find a combination that is 222 using pencil and paper. I reasoned as follows:
1) Goats and pigs both sell for a multiple of $5, but $1661 is not a multiple of 5. Since the price of a cow is an even number, selling 3 cows for $32 each would make the remaining sale $1565.
2) Then the maximum number of animals would be if all sales were 313 pigs. But this is a sale of 316 animals, which is 94 more than 222.
3) This is resolved by selling 94/2 = 47 goats, and 47*3 = 141 fewer pigs.
So, one valid combination of 222 animals is 3 cows, 47 goats, and 172 pigs.
So, 222 must be the solution to the problem.
Edited on June 1, 2022, 1:12 pm
re: Solution
