# Difference between revisions of "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? $\textbf{(A) } 5\qquad \textbf{(B) } 10\qquad \textbf{(C) } 30\qquad \textbf{(D) } 90\qquad \textbf{(E) } 210$

## 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 Lemma tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the GCD (greatest common divisor) of $a$ and $b$. Therefore, the answer is $gcd(300,210)=\boxed{\textbf{(C) }30}.$

## 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 $30$. Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly. Since our supposition can be achieved, the answer is $\boxed{\textbf{(C) }30}$.

## 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 make a non-integer with a linear combination of integers. Since we originally divided by $30$, we need to multiply again, thus getting the answer $1\cdot 30 = \boxed{\textbf{(C) }30}$.

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