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2006 AMC 12A Problems/Problem 14

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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 300 dollars and that goats are worth 210 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 390 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?

\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 30\qquad \mathrm{(D) \ } 90\qquad \mathrm{(E) \ }  210

Solution

The problem can be restated as an equation of the form 300p + 210g = x, where p is the number of pigs, g is the number of goats, and x is the positive debt. The problem asks us to find the lowest x possible. p and g must be integers, which makes the equation a Diophantine equation. The Euclidean algorithm tells us that there are integer solutions to the Diophantine equation am + bn = c, where c is the greatest common divisor of a and b, and no solutions for any smaller c. Therefore, the answer is the greatest common divisor of 300 and 210, which is 30, \mathrm{(C) \ }

Alternatively, note that 300p + 210g = 30(10p + 7g) is divisible by 30 no matter what p and g are, so our answer must be divisible by 30. In addition, three goats minus two pigs give us 630 - 600 = 30 exactly. Since our theoretical best can be achived, it must really be the best, and the answer is \mathrm{(C) \ }. debt that can be resolved.

See also

2006 AMC 12A (ProblemsResources)
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
2006 AMC 10A (ProblemsResources)
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
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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