Answer: Tuesday Teaser #33
This can be solved using the classic trial and error method, but we can also do it mathematically by using the information to set up a couple of simultaneous equations.
Let’s call the number of dogs x, the number of cats y, and the number of mice z.
The first thing we’re told is that there must be 100 animals, so x + y + z = 100.
The second thing is that the 100 animals must cost $100 in total. So 15x + y + 0.25z = 100.
The mathematically-minded will have noticed that this is only two equations for three unknowns. Fortunately, however, there is one other piece of information that we can use - namely that the three unknowns, x, y and z must be whole numbers.
Subtracting the two equations gives: 14x = 0.75z, which simplifies to 56x = 3z.
This is crucial, as we can now deduce that x must be divisible by 3. Given that all dogs cost $15 and the total cost must be $100, this means that there are only two options for x, the number of dogs: 3 or 6. If we were to buy 6 dogs, that would leave only $10 remaining to purchase 94 animals, so we can conclude that x=3.
This immediately allows us to conclude that z=56, from which it easily follows that y=41.
So we must buy 3 dogs, 41 cats and 56 mice.
