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?
  
<math> \mathrm{(A) \ } $5\qquad \mathrm{(B) \ } $10\qquad \mathrm{(C) \ } $30\qquad \mathrm{(D) \ } $90 \mathrm{(E) \ }  $210</math>
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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>
  
 
== Solution ==
 
== Solution ==

Revision as of 00:03, 27 December 2006

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) \ } $210$ (Error compiling LaTeX. Unknown error_msg)

Solution

We see that any amount of debt can be expressed as follows: (300x+210y)-(300a+210b) = m > 0. Rearranging the terms, we have: 300(x-a) + 210(y-b) = m. From here, we note that (x-a) and (y-b) can be any integer. Thus, we let A = (x-a) and B = (y-b). The equation therefore becomes: 300A + 210B = m, where me intend to minimize the value of m. Dividing each side of the equation by the coefficients' GCD, we get the following: 10A + 7B = m/30. Since we know that A and B must be integers, we observe that m/30 must also be an integer. Thus, the smallest possible value m can hold is 30. However, we must check that the equation can be satisfied when m = 30. We can easily see that A = -1 and B = 3 satisfy the equation. Thus, m = 30 is the smallest possible positive debt that can be resolved.

See also