Diophantine Equation [closed]
- by ANIL
In mathematics, a Diophantine equation (named for Diophantus of Alexandria, a third century Greek mathematician) is a polynomial equation where the variables can only take on integer values. Although you may not realize it, you have seen Diophantine equations before: one of the most famous Diophantine equations is:
X^n+Y^n=Z^n
We are not certain that McDonald's knows about Diophantine equations (actually we doubt that they do), but they use them! McDonald's sells Chicken McNuggets in packages of 6, 9 or 20 McNuggets. Thus, it is possible, for example, to buy exactly 15 McNuggets (with one package of 6 and a second package of 9), but it is not possible to buy exactly 16 nuggets, since no non- negative integer combination of 6's, 9's and 20's adds up to 16. To determine if it is possible to buy exactly n McNuggets, one has to solve a Diophantine equation: find non-negative integer values of a, b, and c, such that
6a + 9b + 20c = n.
Problem 1
Show that it is possible to buy exactly 50, 51, 52, 53, 54, and 55 McNuggets, by finding solutions to the Diophantine equation. You can solve this in your head, using paper and pencil, or writing a program. However you chose to solve this problem, list the combinations of 6, 9 and 20 packs of McNuggets you need to buy in order to get each of the exact amounts. Given that it is possible to buy sets of 50, 51, 52, 53, 54 or 55 McNuggets by combinations of 6, 9 and 20 packs, show that it is possible to buy 56, 57,..., 65 McNuggets. In other words, show how, given solutions for 50-55, one can derive solutions for 56-65.
Problem 2
Write an iterative program that finds the largest number of McNuggets that cannot be bought in exact quantity. Your program should print the answer in the following format (where the correct number is provided in place of n):
"Largest number of McNuggets that cannot be bought in exact quantity: n"
Hints:
Hypothesize possible instances of numbers of McNuggets that cannot be purchased exactly, starting with 1
For each possible instance, called n,
a. Test if there exists non-negative integers a, b, and c, such that 6a+9b+20c = n. (This can be done by looking at all feasible combinations of a, b, and c)
b. If not, n cannot be bought in exact quantity, save n
When you have found six consecutive values of n that in fact pass the test of having an exact solution, the last answer that was saved (not the last value of n that had a solution) is the correct answer, since you know by the theorem that any amount larger can also be bought in exact quantity
