2006 AMC 12A Problems/Problem 14
- The following problem is from both the 2006 AMC 12A #14 and 2006 AMC 10A #22, so both problems redirect to this page.
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?
Solutions
Solution 1
The problem can be restated as an equation of the form , where is the number of pigs, is the number of goats, and is the positive debt. The problem asks us to find the lowest x possible. and must be integers, which makes the equation a Diophantine equation. Bezout’s Identity tells us that the smallest for the Diophantine equation to have solutions is when is the greatest common divisor of and . Therefore, the answer is , which is ,
Solution 2
Alternatively, note that is divisible by 30 no matter what and are, so our answer must be divisible by 30. In addition, three goats minus two pigs give us exactly. Since our theoretical best can be achieved, it must really be the best, and the answer is . debt that can be resolved.
Solution 3
Let us simplify this problem. Dividing by , we get a pig to be: , and a goat to be . It becomes evident that if you exchange pigs for goats, we get the smallest positive difference - . Since we originally divided by , we need to multiply again, thus getting the answer:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.