2006 AMC 12A Problems/Problem 14
Problem
Two farmers agree that pigs are worth 
dollars and that goats are worth 
dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a 
dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?
5\qquad \mathrm{(B) \ } 
30\qquad \mathrm{(D) \ } 
210
300p + 210g = x
p
g
x![$is the positive debt. The problem asks us to find the lowest ''x'' possible. ''p'' and ''g'' must be [[integer]]s, which makes the equation a [[Diophantine equation]]. The [[Euclidean algorithm]] tells us that there are integer solutions to the Diophantine equation$](http://latex.artofproblemsolving.com/6/d/8/6d88723ee36acdf891a57a4219c9164cfadfe85e.png)
am + bn = cc![$is the [[greatest common divisor]] of$](http://latex.artofproblemsolving.com/2/b/3/2b3acf5207299c9994ccb9a0800060cbf2649fd6.png)
a
b
c
\mathrm{(C) \ }
300p + 210g = 30(10p + 7g)
pg
630 - 600 = 30
\mathrm{(C) \ }$.
debt that can be resolved.
See also
| 2006 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
