AMC 10 · 2006 · #22
Grade 7 arithmeticTwo 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?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A debt is settled by handing over some pigs (worth $300 each) and some goats (worth $210 each), and change may come back in the other animal. Because change flips the direction of an animal, each kind can effectively be given (counts as positive) or received (counts as negative). We want the smallest positive dollar debt that such an exchange can settle exactly.
Givens: A pig is worth $300 and a goat is worth $210; A debt is paid with pigs or goats, with change returned in goats or pigs as needed; Example: a $390 debt is two pigs given with one goat received as change ($600 - $210 = $390); Answer choices: (A) 5, (B) 10, (C) 30, (D) 90, (E) 210
Unknowns: The smallest positive debt (in dollars) that can be settled exactly this way
Understand
Restated: A debt is settled by handing over some pigs (worth $300 each) and some goats (worth $210 each), and change may come back in the other animal. Because change flips the direction of an animal, each kind can effectively be given (counts as positive) or received (counts as negative). We want the smallest positive dollar debt that such an exchange can settle exactly.
Givens: A pig is worth $300 and a goat is worth $210; A debt is paid with pigs or goats, with change returned in goats or pigs as needed; Example: a $390 debt is two pigs given with one goat received as change ($600 - $210 = $390); Answer choices: (A) 5, (B) 10, (C) 30, (D) 90, (E) 210
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #6 Guess and Check
The pigs and goats have no fixed count, so Tool #4 (Introduce a Variable) names them: let $p$ be the net number of pigs and $g$ the net number of goats, where a negative value means that animal comes back as change. Every settleable debt is then $300p + 210g$. Tool #7 (Identify Subproblems) splits the question into two cleaner pieces: first find a floor no debt can drop below, then check that the floor is actually reachable. The floor falls out of a common factor shared by $300$ and $210$; reaching it is a small Tool #6 (Guess and Check) search for whole numbers, using negatives to model change. A floor that is also achievable is exactly the smallest positive debt.
Execute — Answer: C
6.EE.B.6 Step 1 Model the debt with signed counts
- Let $p$ be the net number of pigs and $g$ the net number of goats that change hands, where a negative count means that animal is received back as change.
- The dollar value settled is $300p + 210g$.
- In the given example, two pigs given and one goat received is $p = 2$, $g = -1$, giving $300(2) + 210(-1) = 390$, matching the $390 debt.
💡 Letting the counts go negative is what turns 'change received' into ordinary subtraction, so one formula covers both giving and getting.
6.NS.B.4 Step 2 Pull out the shared factor for a floor
- Both prices share a common factor.
- The greatest common factor of $300$ and $210$ is $30$, since $300 = 30 \times 10$ and $210 = 30 \times 7$.
- Factoring it out of the debt gives $300p + 210g = 30(10p + 7g)$.
- Since $10p + 7g$ is always a whole number, every settleable debt is a whole-number multiple of $30$.
- So no positive debt can be smaller than $30$.
💡 A common factor of both prices divides every combination of them, so it sets a floor nothing can slip below.
7.NS.A.1 Step 3 Show the floor is reachable
- Now check whether $30$ itself can happen, which needs $10p + 7g = 1$.
- Because $10$ and $7$ share no factor bigger than $1$, such whole numbers exist.
- Testing $p = -2$ and $g = 3$ gives $10(-2) + 7(3) = -20 + 21 = 1$.
- Here $g = 3$ means three goats are given and $p = -2$ means two pigs are received as change: $210(3) - 300(2) = 630 - 600 = 30$.
💡 Since $10$ and $7$ have no common factor, mixing them in whole amounts can land on exactly $1$, and scaling by $30$ lands on the floor itself.
6.NS.B.4 Step 4 Combine floor and reachability
Every possible debt is a multiple of $30$, so nothing positive beats $30$; and the exchange 'give three goats, receive two pigs' settles exactly $30. A reachable floor is the minimum, so the smallest positive debt is $\$30$, which is $\textbf{(C)}$.
💡 A value that nothing can undercut and that can actually occur is by definition the smallest.
6.EE.B.6 Let $p$ be the net number of pigs and $g$ the net number of goats that change ha 6.NS.B.4 Both prices share a common factor. The greatest common factor of $300$ and $210$ 7.NS.A.1 Now check whether $30$ itself can happen, which needs $10p + 7g = 1$. Because $1 6.NS.B.4 Every possible debt is a multiple of $30$, so nothing positive beats $30$; and t Review
Reasonableness: The answer $30$ must divide both prices, and it does: $300 = 30 \times 10$ and $210 = 30 \times 7$. It also fits the given example, since $390 = 30 \times 13$ is a multiple of $30$. The distractors fail the floor test: $5$ (A) and $10$ (B) are smaller than $30$, but no debt can beat the $30$ floor, so they are impossible; $90$ (D) is settleable (it equals $300 - 210$, one pig given and one goat received) but it is $30 \times 3$, not the smallest; $210$ (E) is just one goat's price, far from minimal. Only $30$ is both reachable and unbeatable.
Alternative: Compute the greatest common divisor directly with the Euclidean algorithm: $300 = 1 \times 210 + 90$, then $210 = 2 \times 90 + 30$, then $90 = 3 \times 30 + 0$, so $\gcd(300, 210) = 30$. A standard fact about linear combinations is that the smallest positive value of $300p + 210g$ over integers is exactly this gcd, giving $30$ again.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the net pig count $p$ and goat count $g$ and writing the settled debt as the expression $300p + 210g$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Taking the greatest common factor of $300$ and $210$ as $30$ and factoring $300p + 210g = 30(10p + 7g)$ to prove every debt is a multiple of $30$.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Using a negative count to model animals received as change and combining $210(3) - 300(2) = 30$ to reach the floor.)
⭐ When you can pay and also get change, every amount you can settle is a multiple of the greatest common factor of the two prices, so the smallest debt is that factor itself: here $\gcd(300, 210) = 30$.
⭐ When you can pay and also get change, every amount you can settle is a multiple of the greatest common factor of the two prices, so the smallest debt is that factor itself: here $\gcd(300, 210) = 30$.
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