Difference between revisions of "2006 AMC 12A Problems/Problem 14"
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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 dollars and that goats are worth
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
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 (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 |