AMC 10 · 2020 · #9

Grade 6 algebra

Archetype Small-Domain Constrained Search

perfect-squaresbound-inequality-then-enumerateparity identify-subproblemsbound-inequality-then-enumeratecasework ↑ Prerequisites: perfect-squaresexponents

📏 Medium solution 💡 2 insights

Problem

How many ordered pairs of integers $(x, y)$ satisfy the equation $x^{2020}+y^2=2y?$

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

infinitely many

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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

#7 Identify Subproblems 6.EE.A.3 Step 1

Move the $2y$ to the left and complete the square on the $y$-terms.

$$x^{2020} + y^2 - 2y = 0 \;\Rightarrow\; x^{2020} + (y - 1)^2 = 1$$

💡 Complete the square so both sides are nonnegative.

#9 Solve an Easier Related Problem 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.

$$\{x^{2020}, (y-1)^2\} = \{0, 1\}$$

💡 Nonnegative integers summing to $1$ split as $0 + 1$.

#2 Make a Systematic List 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 $(0, 2)$.

$$x = 0,\quad y \in \{0, 2\} \;\Rightarrow\; (0, 0),\; (0, 2)$$

💡 $0 + 1 = 1$ branch.

#2 Make a Systematic List 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 $(-1, 1)$.

$$x \in \{-1, 1\},\quad y = 1 \;\Rightarrow\; (1, 1),\; (-1, 1)$$

💡 $1 + 0 = 1$ branch.

#7 Identify Subproblems 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: $(0, 0), (0, 2), (1, 1), (-1, 1)$.

$$2 + 2 = 4$$

💡 Disjoint cases sum directly.

#3 Eliminate Possibilities 4.NBT.A.2 Step 6

$4$ matches choice (D).

$$4 \Rightarrow \textbf{(D)}$$

💡 Read the matching answer choice.

[1] #7 6.EE.A.3 Move the $2y$ to the left and complete the square on the $y$-terms.

[2] #9 6.EE.A.1 Both $x^{2020}$ and $(y-1)^2$ are nonnegative integers (the exponent $2020$ is e

[3] #2 6.EE.B.5 Scenario 1: $x^{2020} = 0$ and $(y - 1)^2 = 1$. The first forces $x = 0$. The se

[4] #2 6.EE.B.5 Scenario 2: $x^{2020} = 1$ and $(y - 1)^2 = 0$. The first forces $x = \pm 1$ (on

[5] #7 2.OA.A.1 Add the two scenarios. They are disjoint (different $y$-values), so the total is

[6] #3 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.3 Apply 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.1 Write 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.5 Understand 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.1 Solve one- and two-step word problems using addition and subtraction within 100 (Adding the two scenario counts: $2 + 2 = 4$.)
4.NBT.A.2 Read 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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