Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(numbers)
m (→‎Solution 3)
 
(33 intermediate revisions by 19 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #14]] and [[2006 AMC 10A Problems/Problem 22|2006 AMC 10A #22]]}}
 +
 
== Problem ==
 
== Problem ==
  
 
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\qquad \mathrm{(E) \ }  210</math>
+
<math> \textbf{(A) } 5\qquad \textbf{(B) } 10\qquad \textbf{(C) } 30\qquad \textbf{(D) } 90\qquad \textbf{(E) }  210</math>
 +
 
 +
==Solution 1 (Diophantine Equation)==
 +
 
 +
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. ''<math>p</math>'' and ''<math>g</math>'' must be integers, which makes the equation a [[Diophantine equation]].  [[Bezout%27s Lemma]] tells us that the smallest <math>c</math> for the Diophantine equation <math>am + bn = c</math> to have solutions is when <math>c</math> is the GCD ([[greatest common divisor]]) of <math>a</math> and <math>b</math>. Therefore, the answer is <math>gcd(300,210)=\boxed{\textbf{(C) }30}.</math>
 +
 
 +
==Solution 2 (Divisibility Analysis)==
 +
Alternatively, note that <math>300p + 210g = 30(10p + 7g)</math> is divisible by <math>30</math> no matter what <math>p</math> and <math>g</math> are, so our answer must be divisible by <math>30</math>. Since we want the smallest integer, we can suppose that the answer is <math>30</math> and go on from there. Note that three goats minus two pigs gives us <math>630 - 600 = 30</math> exactly. Since our supposition can be achieved, the answer is <math>\boxed{\textbf{(C) }30}</math>.
  
−
== Solution ==
+
==Solution 3 (Simplifying the Problem)==
−
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>
 
  
−
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>.
+
Let us simplify this problem.  Dividing by <math>30</math>, we get a pig to be:  <math>\frac{300}{30} = 10</math>, and a goat to be <math>\frac{210}{30}= 7</math>.
−
debt that can be resolved.
+
It becomes evident that if you exchange <math>5</math> pigs for <math>7</math> goats, we get the smallest positive difference - <math>5\cdot 10 - 7\cdot 7 = 50-49 = 1</math>, since we can't make a non-integer with a linear combination of integers.   
 +
Since we originally divided by <math>30</math>, we need to multiply again, thus getting the answer <math>1\cdot 30 = \boxed{\textbf{(C) }30}</math>.
  
 
== See also ==
 
== See also ==
 +
Bezout's Lemma:
 +
 +
https://brilliant.org/wiki/bezouts-identity/
 +
 
{{AMC12 box|year=2006|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2006|ab=A|num-b=13|num-a=15}}
 +
{{AMC10 box|year=2006|ab=A|num-b=21|num-a=23}}
 +
{{MAA Notice}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Latest revision as of 02:13, 8 August 2022

The following problem is from both the 2006 AMC 12A #14 and 2006 AMC 10A #22, so both problems redirect to this page.

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?

$\textbf{(A) } 5\qquad \textbf{(B) } 10\qquad \textbf{(C) } 30\qquad \textbf{(D) } 90\qquad \textbf{(E) } 210$

Solution 1 (Diophantine Equation)

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 integers, which makes the equation a Diophantine equation. Bezout's Lemma tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the GCD (greatest common divisor) of $a$ and $b$. Therefore, the answer is $gcd(300,210)=\boxed{\textbf{(C) }30}.$

Solution 2 (Divisibility Analysis)

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 $30$. Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly. Since our supposition can be achieved, the answer is $\boxed{\textbf{(C) }30}$.

Solution 3 (Simplifying the Problem)

Let us simplify this problem. Dividing by $30$, we get a pig to be: $\frac{300}{30} = 10$, and a goat to be $\frac{210}{30}= 7$. It becomes evident that if you exchange $5$ pigs for $7$ goats, we get the smallest positive difference - $5\cdot 10 - 7\cdot 7 = 50-49 = 1$, since we can't make a non-integer with a linear combination of integers. Since we originally divided by $30$, we need to multiply again, thus getting the answer $1\cdot 30 = \boxed{\textbf{(C) }30}$.

See also

Bezout's Lemma:

https://brilliant.org/wiki/bezouts-identity/

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_14&oldid=176774"