Diophantine Equations, Apples, BananasDate: 06/08/97 at 06:28:53 From: Anna Subject: Diophantine Equations Mick has $5.63 worth of bananas and apples. Bananas are 13 cents each and apples are 7 cents each. How many different combinations of apples and bananas can he have and what are they? Date: 07/13/97 at 14:44:43 From: Doctor Terrel Subject: Re: Diophantine Equations Dear Anna, I see that you are interested in a famous math topic called Diophantine Equations. First, we must remember that the solutions to your particular type of problem are always positive integers (i.e. the whole numbers). I mean, it would not do to try to count "negative" apples or bananas. Furthermore, we don't want to have parts of bananas and apples all over the place - you certainly can't do that in a store! This makes our job easier, as I'm sure you can see. Next, we need to set up the equation for our problem. Let a equal the number of apples and b equal the number of bananas we wish to buy. This gives us "7a" and "13b" as the number of cents we will spend for each type of fruit. Since the sum is given in dollars/cents ($5.63), I suggest we simplify matters a little by changing this to just cents, i.e., 563 cents. Now we can write an equation expressing the number of apples and bananas that Mick can buy for 563 cents: 7a + 13b = 563 This should look familiar if you've taken Algebra I in school; it's a linear equation, right? Usually you work with things in x and y, like 3x + 4y = 15. They make straight lines when you graph them; for that reason, they're called linear. Usually you solve for the variable y, in terms of x, like this: 4y = 15 - 3x 15 - 3x y = ---------- 4 Then you make a chart of values, called a T-chart in some books. You choose various values for x, then compute what y will be. It might look like this: x | y ---------|-------- 1 | 3 2 | 2.25 3 | 1.5 4 | 0.75 5 | 0 Of course, in this example the output values (y) do not have to be whole numbers. Decimals (or fractions) are okay; not so in our apple/banana problem. But that's my point - you simply must do the same for 7a + 13b = 563; there are just different numbers in it. But here are two final hints to make your Diophantine work even more efficiently. (1) Solve for the letter that has the smaller number with it, i.e. "a". This means our equation will be: 563 - 13b a = ----------- 7 If a whole number solution exists - and just because you have an equation, there is no guarantee there is one - you will need to try no more than 7 guesses for "b". This is based on the division fact that there are only six possible remainders (1 to 6) when you divide a whole number by 7. Here you should begin substituting 1, 2, 3, ... for "b". If after the 6 you still don't achieve a whole number value for "a", then 7 will do the job! (2) Our T-chart for this problem begins like this: b | a ------|-------- 4 | 73 | Now here comes a beautiful feature that you can discover after trying many more values, in order, for "b". If you increase "b" by 7 (the "a" number in the equation), you get 11, which gives you another whole number output for the "a". This is: a = 60, and we have another solution! Just continue in this manner and you'll find all the positive integer solutions for buying apples and bananas. I hope you can use a calculator to do some of those computations. This sort of problem can be tiring with paper and pencil, but it's fun with a calculator. If you need more help, just let me know. -Doctor Terrel, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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