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.
There is probably an elegant way to solve this with modular arithmetic, but I used a little analytic and then brute force.
The number of animals must be a multiple of both 3 and 37, so it is a multiple of 111.
The average cost per animal is a weighted average of (5,15,32).
1661/111 = 14.96
1661/222 = 7.48
1661/333 = 4.99 which is less than 5, so this cannot be a solution.
So the maximum possible number of animals is 222. But it remains to find some combination of different animals that satisfies the other boundary conditions (or should I call it "fence" conditions?)
I ran this code:
def gain(a,b,c):
cost = [32,15,5]
return cost[0]*a + cost[1]*b + cost[2]*c
start = 1661
gmax = int(start/15)
cmax = int(start/32)
solutions = []
for c in range(cmax+1):
for g in range(gmax,-1,-1):
remain = start - 32*c - 15*g
animals = c+g+remain/5
if remain%5 == 0 and animals%111 == 0:
if remain/5 > 0:
print(c,g,remain/5, gain(c,g,remain/5), ' animals:', animals)
if animals == 222:
solutions.append([c,g,int(remain/5)])
print(solutions)
Output:
3 47 172.0 1661.0 animals: 222.0
8 89 14.0 1661.0 animals: 111.0
13 20 189.0 1661.0 animals: 222.0
18 62 31.0 1661.0 animals: 111.0
28 35 48.0 1661.0 animals: 111.0
38 8 65.0 1661.0 animals: 111.0
[[3, 47, 172], [13, 20, 189]]
Suggesting two possible solutions for 222 animals, $1661 gross income:
cows, goats, pigs:
3, 47, 172
13, 20, 189
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Posted by Larry
on 2022-06-01 07:52:28 |
Solution
