How Many Mice, Cats, and Dogs?Date: 01/13/2003 at 15:32:47 From: Dale Subject: How to express a problem as a formula You are given $100 to buy pets. You must spend all of the money and you must buy 100 pets. You must choose at least one of each pet. The pets and their prices are: mice @ $0.25 each, cats @ $1.00 each, and dogs @ $15.00 each. How many mice, cats, and dogs must you buy? I do not know how to express this in a formula, but I can work it out through trial and error. 56 * $.25 = $14 3 * $15 = $45 41 * $1 = $41 56 + 3 + 41 = 100 $14 + $45 + $41 = $100 Date: 01/14/2003 at 00:40:38 From: Doctor Greenie Subject: Re: How to express a problem as a formula Hello, Dale - Your problem is one involving diophantine equations - sets of equations in which you have more unknowns than equations, but the possible solutions are limited (often, as in this case, to a single solution) by the fact that the values of the unknowns are integers. You can learn more about the general method for solving problems like this by searching the Dr. Math archives using the keyword diophantine Here in particular are links to two pages where you can find detailed discussions of a problem similar to yours: Buying Cows, Pigs, and Chickens http://mathforum.org/library/drmath/view/53051.html Farmer Buying Livestock http://mathforum.org/library/drmath/view/57395.html The numbers in your problem lend themselves very nicely to understanding the general method for solving this type of problem, so let's look at the solution in your case. We start with m = number of mice c = number of cats d = number of dogs Then the total number of animals is 100: m+c+d = 100 And the total cost is $100: (m/4)+c+15d = 100 We have used all the given information, but we still have three unknowns and only two equations. We subtract these two equations to get 14d - (3/4)m = 0 14d = (3/4)m 56d = 3m Now we use the fact that d and m are both integers... m = (56/3)d The fraction 56/3 is not an integer; for m to be an integer, d must be a multiple of 3. Trying d=0 is not allowed, because the problem requires that you choose at least one of each type of pet. Trying d=3 gives us m=56, which leads to c=41, which is the answer you found. Trying d=6 gives m=112, which is impossible because the total number of animals is 100. So the answer you found is the only one. I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/