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

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<math> \textbf{(A) } 5\qquad \textbf{(B) } 10\qquad \textbf{(C) } 30\qquad \textbf{(D) } 90\qquad \textbf{(E) }  210</math>
 
<math> \textbf{(A) } 5\qquad \textbf{(B) } 10\qquad \textbf{(C) } 30\qquad \textbf{(D) } 90\qquad \textbf{(E) }  210</math>
  
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==Solution 1==
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==Solution 1 (Diophantine Equation)==
  
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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. ''<math>p</math>'' and ''<math>g</math>'' must be integers, which makes the equation a [[Diophantine equation]].  Bezout’s Identity 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>
+
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>
  
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==Solution 2==
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==Solution 2 (Divisibility Analysis)==
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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>. In addition, three goats minus two pigs gives us <math>630 - 600 = 30</math> exactly.  Since our theoretical best can be achieved, it must really be the best, and the answer is <math>\mathrm{(C) \ }</math>.
+
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>.
  
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==Solution 3==
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==Solution 3 (Simplifying the Problem)==
  
 
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>.  
 
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>.  
 
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.   
 
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.   
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Since we originally divided by <math>30</math>, we need to multiply again, thus getting the answer: <math>1\cdot 30 = \mathrm{(C) 30}</math>
+
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 Identity:
+
Bezout's Lemma:
−
 
 
−
https://artofproblemsolving.com/wiki/index.php/Bezout%27s_Lemma
 
  
 
https://brilliant.org/wiki/bezouts-identity/
 
https://brilliant.org/wiki/bezouts-identity/

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

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