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 $300$ 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 1

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. Bezout’s Identity tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the greatest common divisor of $a$ and $b$ . Therefore, the answer is $gcd(300,210)$ , which is $30$ , $\mathrm{(C) \ }$

Solution 2

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 gives us $630 - 600 = 30$ exactly. Since our theoretical best can be achieved, it must really be the best, and the answer is $\mathrm{(C) \ }$ .

Solution 3

Let us simplify this problem. Dividing by $30$ , we get a pig to be: $\frac{300}{30} = 10$ , and a goat to be $\frac{210}{30}= 7$ . It becomes evident that if you exchange $5$ pigs for $7$ goats, we get the smallest positive difference - $5\cdot 10 - 7\cdot 7 = 50-49 = 1$ , since we can't made a non-integer with a linear combination of integers. Since we originally divided by $30$ , we need to multiply again, thus getting the answer: $1\cdot 30 = \mathrm{(C) 30}$

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2006 AMC 12A Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Solution 3

See also