AMC 10 · 2012 · #24
Grade 8 arithmeticLet a, b, and c be positive integers with a≥ b≥ c such that
a2−b2−c2+ab=2011 and
a2+3b2+3c2−3ab−2ac−2bc=−1997.
What is a?
Pick an answer.
AMC 10 2012 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: Positive integers $a \ge b \ge c$ satisfy two messy equations: $a^2 - b^2 - c^2 + ab = 2011$ and $a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997$. Find the value of $a$.
Givens: $a$, $b$, $c$ are positive integers; $a \ge b \ge c$; Equation (1): $a^2 - b^2 - c^2 + ab = 2011$; Equation (2): $a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997$; Answer choices: (A) 249, (B) 250, (C) 251, (D) 252, (E) 253
Unknowns: The value of $a$
Understand
Restated: Positive integers $a \ge b \ge c$ satisfy two messy equations: $a^2 - b^2 - c^2 + ab = 2011$ and $a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997$. Find the value of $a$.
Givens: $a$, $b$, $c$ are positive integers; $a \ge b \ge c$; Equation (1): $a^2 - b^2 - c^2 + ab = 2011$; Equation (2): $a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997$; Answer choices: (A) 249, (B) 250, (C) 251, (D) 252, (E) 253
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities, #7 Identify Subproblems
Each equation alone is a tangle of squares and cross-terms. Tool #15 (Organize Information in More Ways) is the key move: add the two equations. The nasty cross-terms line up so the sum collapses into the symmetric shape $a^2 + b^2 + c^2 - ab - bc - ca$, which is exactly half of $(a-b)^2 + (b-c)^2 + (c-a)^2$. That turns the problem into a statement about the gaps between the three numbers. Tool #4 (Introduce a Variable) names those gaps $x = a-b$ and $y = b-c$; Tool #3 (Eliminate Possibilities) pins the gaps down because a small sum of squares has only a few whole-number solutions; Tool #7 (Identify Subproblems) then splits the finish into two cases and checks each against equation (1).
Execute — Answer: E
8.EE.C.8 Step 1 Add the two equations
- Add equation (1) and equation (2) term by term.
- The $a^2$ terms give $2a^2$; the $b^2$ terms give $-b^2 + 3b^2 = 2b^2$; the $c^2$ terms give $-c^2 + 3c^2 = 2c^2$; the $ab$ terms give $ab - 3ab = -2ab$; and $-2ac$, $-2bc$ carry over.
- The right side becomes $2011 + (-1997) = 14$.
- Divide everything by $2$.
💡 Two equations about the same unknowns can be added; here the ugly cross-terms partly cancel and leave a clean symmetric expression.
6.EE.A.3 Step 2 Rewrite as a sum of squares
- The symmetric expression $a^2 + b^2 + c^2 - ab - bc - ca$ is exactly half of $(a-b)^2 + (b-c)^2 + (c-a)^2$ (expand the three squares and you get $2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca$).
- So multiply both sides by $2$.
💡 A symmetric quadratic in three variables repackages into squared differences, which measure how far apart the numbers are.
6.EE.B.6 Step 3 Name the gaps
- Let $x = a - b$ and $y = b - c$.
- Because $a \ge b \ge c$, both $x$ and $y$ are non-negative integers, and the largest difference is $a - c = x + y$.
- Substitute into the squared-difference equation: $x^2 + y^2 + (x+y)^2 = 14$.
- Expand $(x+y)^2$ and combine to get $2x^2 + 2y^2 + 2xy = 14$, then divide by $2$.
💡 Working with the gaps instead of the numbers themselves shrinks three unknowns down to two small ones.
6.EE.B.5 Step 4 Pin down the gaps
- Search non-negative integers $x, y$ with $x^2 + xy + y^2 = 7$.
- Each variable must be at most $2$ (if $x \ge 3$ then $x^2 \ge 9 > 7$).
- Testing $x = 1, y = 2$: $1 + 2 + 4 = 7$.
- Testing $x = 2, y = 1$: $4 + 2 + 1 = 7$.
- No other pair works ($x = y$ gives $3x^2 = 7$, impossible).
- So the gaps are $\{x, y\} = \{1, 2\}$ in some order, and either way $a - c = x + y = 3$.
💡 A sum of squares equal to a small number has only a handful of whole-number solutions, so you can just list them.
6.EE.A.1 Step 5 Turn each case into an equation in $a$
- Now use equation (1): $a^2 + ab - b^2 - c^2 = 2011$.
- Write $b$ and $c$ in terms of $a$ for each case and expand the squares.
- Case A ($x=1, y=2$): $b = a-1$, $c = a-3$, which gives $a^2 + a(a-1) - (a-1)^2 - (a-3)^2 = 7a - 10$.
- Case B ($x=2, y=1$): $b = a-2$, $c = a-3$, which gives $a^2 + a(a-2) - (a-2)^2 - (a-3)^2 = 8a - 13$.
💡 Substituting the known gaps collapses the quadratic in (1) into a single linear equation in $a$.
7.NS.A.2 Step 6 Reject Case A
- Case A requires $7a - 10 = 2011$, so $7a = 2021$.
- But $2021 = 7 \times 288 + 5$, which is not a multiple of $7$, so $a$ would not be a whole number.
- Since $a$ must be a positive integer, Case A is impossible.
💡 An integer unknown can only survive if the arithmetic divides evenly; a leftover remainder kills the case.
8.EE.C.7 Step 7 Solve Case B
- Case B requires $8a - 13 = 2011$, so $8a = 2024$ and $a = 253$.
- Then $b = 251$ and $c = 250$, which satisfy $a \ge b \ge c$ and are all positive integers.
- A quick check confirms both original equations hold: $253^2 - 251^2 - 250^2 + 253 \cdot 251 = 2011$ and $253^2 + 3 \cdot 251^2 + 3 \cdot 250^2 - 3 \cdot 253 \cdot 251 - 2 \cdot 253 \cdot 250 - 2 \cdot 251 \cdot 250 = -1997$.
- So $a = 253$, which is answer (E).
💡 The one surviving case gives a clean linear equation whose whole-number solution is the answer.
8.EE.C.8 Add equation (1) and equation (2) term by term. The $a^2$ terms give $2a^2$; the 6.EE.A.3 The symmetric expression $a^2 + b^2 + c^2 - ab - bc - ca$ is exactly half of $(a 6.EE.B.6 Let $x = a - b$ and $y = b - c$. Because $a \ge b \ge c$, both $x$ and $y$ are n 6.EE.B.5 Search non-negative integers $x, y$ with $x^2 + xy + y^2 = 7$. Each variable mus 6.EE.A.1 Now use equation (1): $a^2 + ab - b^2 - c^2 = 2011$. Write $b$ and $c$ in terms 7.NS.A.2 Case A requires $7a - 10 = 2011$, so $7a = 2021$. But $2021 = 7 \times 288 + 5$, 8.EE.C.7 Case B requires $8a - 13 = 2011$, so $8a = 2024$ and $a = 253$. Then $b = 251$ a Review
Reasonableness: Plug $a = 253$, $b = 251$, $c = 250$ back into both givens: equation (1) gives $64009 - 63001 - 62500 + 63503 = 2011$, and equation (2) gives $64009 + 189003 + 187500 - 190509 - 126500 - 125500 = -1997$. Both match exactly, and $253 \ge 251 \ge 250$ respects $a \ge b \ge c$. The gaps $(a-b, b-c) = (2, 1)$ agree with the $\{1,2\}$ solution found from the sum of squares. The answer $253$ is choice (E).
Alternative: Instead of splitting into cases at the end, subtract the two case-equations first: $8a - 13$ versus $7a - 10$ both equal $2011$ only if the corresponding gap layout is consistent, so you can test divisibility right away — $2011 + 10 = 2021$ is not divisible by $7$, but $2011 + 13 = 2024$ is divisible by $8$. That immediately selects Case B and the same $a = 253$, without ever computing $b$ and $c$ separately.
CCSS standards used (min grade 8)
8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Adding the two given equations term by term to combine them into a single simpler equation.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Rewriting the symmetric expression $a^2 + b^2 + c^2 - ab - bc - ca$ as half of $(a-b)^2 + (b-c)^2 + (c-a)^2$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Introducing gap variables $x = a-b$ and $y = b-c$ to reduce three unknowns to two.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Listing the non-negative integer solutions of $x^2 + xy + y^2 = 7$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Expanding the squares $(a-1)^2$, $(a-2)^2$, $(a-3)^2$ when substituting each case into equation (1).)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Testing divisibility of $2021$ by $7$ to reject the case that gives no integer $a$.)8.EE.C.7Solve linear equations in one variable (Solving $8a - 13 = 2011$ to get $a = 253$.)
⭐ When two equations both look ugly, add them first — here the mess collapses into a sum of squared gaps, and once you know the numbers are only $1$, $2$, and $3$ apart, one quick divisibility check leaves $a = 253$.
⭐ When two equations both look ugly, add them first — here the mess collapses into a sum of squared gaps, and once you know the numbers are only $1$, $2$, and $3$ apart, one quick divisibility check leaves $a = 253$.
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