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| − | == Problem ==
| + | #redirect [[2006 AMC 12A Problems/Problem 14]] |
| − | Two farmers agree that pigs are worth $300 and that goats are worth $210. 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 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?
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| − | <math>\mathrm{(A) \ } \$5\qquad\mathrm{(B) \ } \$10\qquad\mathrm{(C) \ } \$30\qquad\mathrm{(D) \ } \$90\qquad\mathrm{(E) \ } \$210\qquad</math>
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| − | == Solution ==
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| − | The problem can be restated as an equation of the form 300''p'' + 210''g'' = ''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 [[integer]]s, which makes the equation a [[Diophantine equation]]. From the [[Euclidean Algorithm]] tells us that there are integer solutions to the Diophantine equation $ax + by = c$, where $c$ is the greatest common denominator of $a$ and $b$, and no solutions for any smaller $c$. Therefore, the answer is simply the [[greatest common divisor]] of 300 and 210, which is 30 $\mathrm{(C) \ }$
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| − | 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 $\mathrm{(C) \ }$.
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| − | == See also ==
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| − | *[[2006 AMC 10A Problems]]
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