AMC 10 · 2020 · #9
Grade 6 algebraProblem
How many ordered pairs of integers (x,y) satisfy the equation x2020+y2=2y?
Pick an answer.
AMC 10 2020 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: Find the number of ordered pairs of integers $(x, y)$ satisfying $x^{2020} + y^2 = 2y$.
Givens: The equation $x^{2020} + y^2 = 2y$; Both $x$ and $y$ are integers; $2020$ is an even positive exponent; Choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) infinitely many
Unknowns: The count of ordered integer pairs $(x, y)$
Understand
Restated: Find the number of ordered pairs of integers $(x, y)$ satisfying $x^{2020} + y^2 = 2y$.
Givens: The equation $x^{2020} + y^2 = 2y$; Both $x$ and $y$ are integers; $2020$ is an even positive exponent; Choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) infinitely many
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #2 Make a Systematic List, #3 Eliminate Possibilities
Tool #7 reshapes the equation by completing the square: $x^{2020} + (y-1)^2 = 1$ — now both terms are nonnegative integers summing to $1$. Tool #9 (easier) replaces the scary $x^{2020}$ with the simpler observation "it's $0$ when $x = 0$ and at least $1$ otherwise (and exactly $1$ only when $x = \pm 1$)". Tool #2 then lists the two cases that achieve the sum $1$. Tool #3 confirms against the choices.
Execute — Answer: D
6.EE.A.3 Step 1 Move the $2y$ to the left and complete the square on the $y$-terms.
💡 Complete the square so both sides are nonnegative.
6.EE.A.1 Step 2 - Both $x^{2020}$ and $(y-1)^2$ are nonnegative integers (the exponent $2020$ is even, and $(y-1)^2$ is a perfect square).
- Two nonnegative integers add to $1$ only when one is $0$ and the other is $1$.
- So there are exactly two scenarios to check.
💡 Nonnegative integers summing to $1$ split as $0 + 1$.
6.EE.B.5 Step 3 - Scenario 1: $x^{2020} = 0$ and $(y - 1)^2 = 1$.
- The first forces $x = 0$.
- The second gives $y - 1 = \pm 1$, so $y = 0$ or $y = 2$.
- That is $2$ pairs: $(0, 0)$ and $(0, 2)$.
💡 $0 + 1 = 1$ branch.
6.EE.B.5 Step 4 - Scenario 2: $x^{2020} = 1$ and $(y - 1)^2 = 0$.
- The first forces $x = \pm 1$ (only integers whose $2020$-th power is $1$).
- The second gives $y = 1$.
- That is $2$ more pairs: $(1, 1)$ and $(-1, 1)$.
💡 $1 + 0 = 1$ branch.
2.OA.A.1 Step 5 - Add the two scenarios.
- They are disjoint (different $y$-values), so the total is $2 + 2 = 4$ ordered pairs: $(0, 0), (0, 2), (1, 1), (-1, 1)$.
💡 Disjoint cases sum directly.
4.NBT.A.2 Step 6 $4$ matches choice (D).
💡 Read the matching answer choice.
6.EE.A.3 Move the $2y$ to the left and complete the square on the $y$-terms. 6.EE.A.1 Both $x^{2020}$ and $(y-1)^2$ are nonnegative integers (the exponent $2020$ is e 6.EE.B.5 Scenario 1: $x^{2020} = 0$ and $(y - 1)^2 = 1$. The first forces $x = 0$. The se 6.EE.B.5 Scenario 2: $x^{2020} = 1$ and $(y - 1)^2 = 0$. The first forces $x = \pm 1$ (on 2.OA.A.1 Add the two scenarios. They are disjoint (different $y$-values), so the total is 4.NBT.A.2 $4$ matches choice (D). Review
Reasonableness: Verify each of the $4$ pairs satisfies the original $x^{2020} + y^2 = 2y$: $(0, 0): 0 + 0 = 0 = 2 \cdot 0$ ✓; $(0, 2): 0 + 4 = 4 = 2 \cdot 2$ ✓; $(1, 1): 1 + 1 = 2 = 2 \cdot 1$ ✓; $(-1, 1): 1 + 1 = 2 = 2 \cdot 1$ ✓. Also confirm no others: any $|x| \geq 2$ makes $x^{2020} \geq 2^{2020}$, vastly larger than $1$, so $(y-1)^2$ would have to be negative — impossible. Choice (E) infinitely many is ruled out.
Alternative: Tool #6 (Guess & Check) on small $x$: try $x = 0, \pm 1, \pm 2, \ldots$ in the original equation and solve for $y$ each time. $x = 0: y^2 = 2y \Rightarrow y(y-2) = 0 \Rightarrow y \in \{0, 2\}$. $x = \pm 1: 1 + y^2 = 2y \Rightarrow (y-1)^2 = 0 \Rightarrow y = 1$. $|x| \geq 2: x^{2020}$ is enormous, no integer $y$ works. Same $4$ pairs.
CCSS standards used (min grade 6)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Rewriting $y^2 - 2y$ as $(y - 1)^2 - 1$ (completing the square) to make the right side a sum of nonnegative pieces equal to $1$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Recognizing that an even exponent makes $x^{2020} \geq 0$, and that $x^{2020} = 1$ only when $x = \pm 1$ over the integers.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Solving $(y - 1)^2 = 1$ and $(y - 1)^2 = 0$ for $y$, and reading off the matching $x$-values from $x^{2020} = 0$ or $1$.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Adding the two scenario counts: $2 + 2 = 4$.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching the count $4$ to choice (D).)
⭐ This AMC 10 problem only needs Grade 6 expression rewriting you already know — complete the square to get $x^{2020} + (y-1)^2 = 1$, then notice two nonnegative whole numbers add to $1$ only as $0 + 1$ or $1 + 0$. That gives $(0, 0), (0, 2), (1, 1), (-1, 1)$ — $4$ pairs.
⭐ This AMC 10 problem only needs Grade 6 expression rewriting you already know — complete the square to get $x^{2020} + (y-1)^2 = 1$, then notice two nonnegative whole numbers add to $1$ only as $0 + 1$ or $1 + 0$. That gives $(0, 0), (0, 2), (1, 1), (-1, 1)$ — $4$ pairs.
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