2006 AMC 12A Problems/Problem 14

Revision as of 14:36, 17 September 2017 by Cocohearts (talk | contribs) (Solution 3)
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?

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

Solutions

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. 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, $\mathrm{(C) \ }$

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

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 originally divided by $30$, we need to multiply again, thus getting the answer: $1\cdot 30 = \mathrm{(C) 30}$

Solution 4

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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