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Difference between revisions of "2006 AMC 12A Problems/Problem 14"

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Two farmers agree that pigs are worth <math>300</math> dollars and that goats are worth <math>210</math> 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 <math>390</math> 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?
 
Two farmers agree that pigs are worth <math>300</math> dollars and that goats are worth <math>210</math> 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 <math>390</math> 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?
  
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<math> \mathrm{(A) \ } $5\qquad \mathrm{(B) \ } $10\qquad \mathrm{(C) \ } $30\qquad \mathrm{(D) \ } $90\qquad \mathrm{(E) \ }  $210</math>
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<math> \mathrm{(A) \ } </math>5\qquad \mathrm{(B) \ } <math>10\qquad \mathrm{(C) \ } </math>30\qquad \mathrm{(D) \ } <math>90\qquad \mathrm{(E) \ }  </math>210<math>
  
 
== Solution ==
 
== Solution ==
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The problem can be restated as an equation of the form <math>300p + 210g = x</math>, where <math>p</math> is the number of pigs, <math>g</math> is the number of goats, and <math>x</math> 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]].  The [[Euclidean algorithm]] tells us that there are integer solutions to the Diophantine equation <math>am + bn = c</math>, where <math>c</math> is the [[greatest common divisor]] of <math>a</math> and <math>b</math>, and no solutions for any smaller <math>c</math>. Therefore, the answer is the greatest common divisor of 300 and 210, which is 30, <math>\mathrm{(C) \ }</math>
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The problem can be restated as an equation of the form </math>300p + 210g = x<math>, where </math>p<math> is the number of pigs, </math>g<math> is the number of goats, and </math>x<math> 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]].  The [[Euclidean algorithm]] tells us that there are integer solutions to the Diophantine equation </math>am + bn = c<math>, where </math>c<math> is the [[greatest common divisor]] of </math>a<math> and </math>b<math>, and no solutions for any smaller </math>c<math>. Therefore, the answer is the greatest common divisor of 300 and 210, which is 30, </math>\mathrm{(C) \ }<math>
  
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Alternatively, note that <math>300p + 210g = 30(10p + 7g)</math> is divisible by 30 no matter what <math>p</math> and <math>g</math> are, so our answer must be divisible by 30.  In addition, three goats minus two pigs give us <math>630 - 600 = 30</math> exactly.  Since our theoretical best can be achived, it must really be the best, and the answer is <math>\mathrm{(C) \ }</math>.
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Alternatively, note that </math>300p + 210g = 30(10p + 7g)<math> is divisible by 30 no matter what </math>p<math> and </math>g<math> are, so our answer must be divisible by 30.  In addition, three goats minus two pigs give us </math>630 - 600 = 30<math> exactly.  Since our theoretical best can be achived, it must really be the best, and the answer is </math>\mathrm{(C) \ }$.
 
debt that can be resolved.
 
debt that can be resolved.
  
 
== See also ==
 
== See also ==
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* [[2006 AMC 12A Problems]]
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{{AMC12 box|year=2006|ab=A|num-b=13|num-a=15}}
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*[[2006 AMC 10A Problems/Problem 22]]
 
  
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{{AMC12 box|year=2006|ab=A|num-b=13|num-a=15}}
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{{AMC10 box|year=2006|ab=A|num-b=21|num-a=23}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Revision as of 16:43, 28 October 2007

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 [[integer]]s, 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 (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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