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

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* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
  
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[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Revision as of 19:20, 2 February 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) \ } $210$ (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 (positive due to the constraint of the problem). 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

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