Dogs, Cats and Mice problem

[Andrzejek]

Here is one that I love. My co-workers were trying to solve it while driving home. Very dangerous. And it is perfect if you want to teach loops in computer programming.

You're given a hundred dollars and told to spend it all purchasing exactly a hundred animals at the pet store. Dogs cost $15. Cats cost a buck, and mice are 25 cents each. And you have to purchase at least one of each animal.

How many Dogs, Cats and Mice do you need to buy?

Have fun.

---- Andy

 

[jges]

Spoiler

3 dogs, 41 cats, and 56 mice

 

[Noway2]

Spoiler

There are multiple solutions. In addition to the above: 6 dogs, 9 cats, 4 mice would work for a minimal solution.

 

[jges]

Noway2,
Your solution does not add up to 100 animals.

 

[CajunCenturion]

[hide]Without using loops:
You have three unknowns and two equations:
x + y + z = 100
15x + y + 0.25z = 100
What you do know is that 1 <= x <= 6, so you only have six possible solutions, each which can be checked with two simultaneous equations:
for x = 1: y + z = 99 and y + 0.25z = 85
for x = 2: y + z = 98 and y + 0.25z = 70
for x = 3: y + z = 97 and y + 0.25z = 55
for x = 4: y + z = 96 and y + 0.25z = 40
for x = 5: y + z = 95 and y + 0.25z = 25
for x = 6: y + z = 94 and y + 0.25z = 10
But, before you do that, you also know that mod(z, 4) = 0, so y + z must be a multiple of 4 plus 1. That eliminates options except for x = 3.
Therefore x = 3, y + z = 97 and y + 0.25z = 55
Solving simultaneously, we get y = 97 - z, so
97 - z + .25z = 55
97 - .75z = 55
.75x = 42
z = 56

y + 56 = 97
y = 41

x = 3 dogs
y = 41 cats
z = 56 mice
[/hide]

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[kwbMitel]

I got Jgres solution and agree that it is the only solution.

*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.

 

[jges]

Excellent explanation by CajunCenturion.

 

[lionelhill]

Spoiler

another way to look at it without loops:
Using the same two equations as Cajun used
(1) 15d+c+(m/4)=100
(2) 15d+15c+15m=100*15
subtract (1) from (2)
14c+(59/4)m=1400
therefore c+(59/56)m=100
Obviously both (59/56)m and m itself must be integers. m is an integer by definition, and (59/56)m must be because otherwise c won't be.
The smallest (and only) solution is m=56, so (59/56)m=59
Therefore c=41
And since the animals add up to 100, d=3

 

[Olaf Doschke]

The not so clever brute force solution of loops has one advantage over all analytical solutions. You can much easier introduce other numbers, eg prices and find their solution too.

#include <stdio.h>
#include <iostream>

#define CAPITAL 200
#define NUMBEROFANIMALS 200

#define DOGPRICE 15
#define CATPRICE 1
#define MOUSEPRICE .25

int main(int argc, char **argv)
{
 int Dogs,Cats,Mice;

 for (Dogs = 1;Dogs <= int (CAPITAL/DOGPRICE); Dogs++)
   for (Cats = 1;Cats <= int ((CAPITAL-DOGPRICE*Dogs)/CATPRICE); Cats++)
      {
       Mice = NUMBEROFANIMALS-Cats-Dogs;
      if (Mice>0 && DOGPRICE*Dogs+CATPRICE*Cats+MOUSEPRICE*Mice == CAPITAL)
         cout << "Solution:", Dogs, Cats, Mice;
      }
 return 0;
}

Here for example both the capital and the number of animals wanted is doubled to 200. There are 3 solutions for that.

Bye, Olaf.

 

[Olaf Doschke]

You can play around with it here:

http://codepad.org/hiqS7gLp

Bye, Olaf.

 

[joel76]

Prolog ise usefull for this kind of game :

:- use_module(library(clpfd)).

dogs_cats_mices(D, C, M) :-
	S = [D, C, M],
	S ins 1..100,
	% Number of animals
	sum(S, #=, 100),
	% Price
	% with clpfd, we can only use integers
	(D * 1500 + C * 100 + M * 25) #= 10000,
	labeling([], S).

And we get :

?- dogs_cats_mices(Dogs, Cats, Mices).
Dogs = 3,
Cats = 41,
Mices = 56 ;
false.