AMC 10 · 2022 · #21

Grade 8 number-theory

polynomial-factoringpolynomial-rootssystems-of-equationspolynomial-remainderconvert-to-algebra easier-related-problemconvert-to-algebraidentify-subproblems ↑ Prerequisites: polynomial-factoringsystems-of-equations

📏 Long solution 💡 4 insights

Problem

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial
$x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 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.

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Toolkit + CCSS Solution

Understand

Restated: Find a polynomial $P(x)$ with rational coefficients that leaves remainder $x+2$ when divided by $x^2+x+1$ and remainder $2x+1$ when divided by $x^2+1$. Among all such polynomials, pick the unique one with the smallest possible degree, then compute the sum of the squares of its coefficients.

Givens: $P(x) = (x^2+x+1)\,Q_1(x) + (x+2)$ for some polynomial $Q_1(x)$; $P(x) = (x^2+1)\,Q_2(x) + (2x+1)$ for some polynomial $Q_2(x)$; $P(x)$ has rational coefficients; Answer choices: (A) $10$, (B) $13$, (C) $19$, (D) $20$, (E) $23$

Unknowns: Sum of squares of the coefficients of the least-degree $P(x)$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #13 Convert to Algebra, #7 Identify Subproblems, #3 Eliminate Possibilities

Tool #9 (Easier Problem) — first try the smallest possible degree for $P$ (degree $2$, with constant quotients) to see what goes wrong; then bump up to degree $3$ (linear quotients). Tool #7 (Subproblems) — splitting equation $A = B$ into three coefficient-equations (one per power of $x$) turns the polynomial puzzle into a small system. Tool #13 (Algebra) — solve that tiny system for the unknown coefficients. Tool #3 (Eliminate) — quickly verify the final sum $1^2+2^2+3^2+3^2 = 23$ matches choice (E).

Execute — Answer: E

#9 Solve an Easier Related Problem 6.EE.A.4 Step 1

Try the easiest case first: what if $P$ has degree $2$?
Then both quotients $Q_1, Q_2$ are constants, say $c_1$ and $c_2$.
Expand: $P = c_1 x^2 + (c_1+1) x + (c_1+2)$ and $P = c_2 x^2 + 2x + (c_2+1)$.
Match $x^2$: $c_1 = c_2$.
Match $x$: $c_1 + 1 = 2$, so $c_1 = 1$.
Match constant: $c_1 + 2 = c_2 + 1$, i.e.
$3 = 2$.
Contradiction!
So degree $2$ is impossible.

$$\text{degree } 2 \;\Rightarrow\; 3 = 2 \;\text{(impossible)}$$

💡 Grade 6 — test if two expressions can be the same expression by matching coefficient terms.

#9 Solve an Easier Related Problem 6.EE.A.2 Step 2

Try degree $3$.
The quotients $Q_1, Q_2$ are now linear.
Both expressions must share the same leading $x^3$ coefficient, so write $Q_1(x) = ax + b$ and $Q_2(x) = ax + c$ (same $a$).

$$Q_1 = ax+b, \quad Q_2 = ax+c$$

💡 Grade 6 — let letters stand for the unknown coefficients.

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

Expand expression 1: $(x^2+x+1)(ax+b) + (x+2) = ax^3 + (a+b)x^2 + (a+b)x + b + x + 2 = ax^3 + (a+b)x^2 + (a+b+1)x + (b+2)$.

$$P = ax^3 + (a+b)x^2 + (a+b+1)x + (b+2)$$

💡 Grade 6 — distribute and collect like terms to get a clean form.

#7 Identify Subproblems 6.EE.A.3 Step 4

Expand expression 2: $(x^2+1)(ax+c) + (2x+1) = ax^3 + cx^2 + ax + c + 2x + 1 = ax^3 + cx^2 + (a+2)x + (c+1)$.

$$P = ax^3 + cx^2 + (a+2)x + (c+1)$$

💡 Grade 6 — same expansion technique, applied to the second form.

#13 Convert to Algebra 8.EE.C.8 Step 5

Match coefficients term-by-term.
$x^3$: both give $a$, automatic.
$x^2$: $a+b = c$.
$x$: $a+b+1 = a+2$, so $b = 1$.
Constant: $b+2 = c+1$, so $3 = c+1$, giving $c = 2$.
Back to $x^2$: $a+1 = 2$, so $a = 1$.

$$a = 1, \quad b = 1, \quad c = 2$$

💡 Grade 8 — three linear equations in three unknowns; solve by substitution.

#13 Convert to Algebra 6.EE.A.2 Step 6

Plug $a=1, b=1$ into expression 1: $P(x) = x^3 + (1+1)x^2 + (1+1+1)x + (1+2) = x^3 + 2x^2 + 3x + 3$.
Quick check with expression 2 using $c = 2$: $x^3 + 2x^2 + (1+2)x + (2+1) = x^3 + 2x^2 + 3x + 3$.
Match.

$$P(x) = x^3 + 2x^2 + 3x + 3$$

💡 Grade 6 — substitute back to read off $P(x)$ and confirm both forms agree.

#3 Eliminate Possibilities 6.EE.A.1 Step 7

Sum of the squares of the coefficients of $P(x) = x^3 + 2x^2 + 3x + 3$: coefficients are $1, 2, 3, 3$.
Compute $1^2 + 2^2 + 3^2 + 3^2 = 1 + 4 + 9 + 9 = 23$.
Answer is (E).

$$1^2 + 2^2 + 3^2 + 3^2 = 23 \;\Rightarrow\; \textbf{(E)}$$

💡 Grade 6 — evaluate a numerical expression with exponents, then check the answer choices.

[1] #9 6.EE.A.4 Try the easiest case first: what if $P$ has degree $2$? Then both quotients $Q_1

[2] #9 6.EE.A.2 Try degree $3$. The quotients $Q_1, Q_2$ are now linear. Both expressions must s

[3] #7 6.EE.A.3 Expand expression 1: $(x^2+x+1)(ax+b) + (x+2) = ax^3 + (a+b)x^2 + (a+b)x + b + x

[4] #7 6.EE.A.3 Expand expression 2: $(x^2+1)(ax+c) + (2x+1) = ax^3 + cx^2 + ax + c + 2x + 1 = a

[5] #13 8.EE.C.8 Match coefficients term-by-term. $x^3$: both give $a$, automatic. $x^2$: $a+b =

[6] #13 6.EE.A.2 Plug $a=1, b=1$ into expression 1: $P(x) = x^3 + (1+1)x^2 + (1+1+1)x + (1+2) = x

[7] #3 6.EE.A.1 Sum of the squares of the coefficients of $P(x) = x^3 + 2x^2 + 3x + 3$: coeffici

Review

Reasonableness: Sanity. Verify $P(x) = x^3 + 2x^2 + 3x + 3$ in both original conditions. Divide by $x^2 + x + 1$: $P = (x^2+x+1)(x+1) + (x+2)$, remainder $x+2$ — matches. Divide by $x^2 + 1$: $P = (x^2+1)(x+2) + (2x+1)$, remainder $2x+1$ — matches. Sum of squared coefficients $1+4+9+9 = 23$, which is exactly choice (E). The degree-$2$ attempt failed by a numerical contradiction ($3 = 2$), so degree $3$ is genuinely the least. The choices $10, 13, 19, 20, 23$ are decoys for forgetting a coefficient or for the unsquared sum $1+2+3+3 = 9$.

Alternative: Tool #3 (Eliminate). The two remainders $x+2$ and $2x+1$ both have degree $\le 1$. By a standard Chinese-remainder-style argument for polynomials, the unique least-degree solution has degree $< \deg((x^2+x+1)(x^2+1)) = 4$, so $\deg P \le 3$. Combined with the degree-$2$ contradiction, $\deg P = 3$ is forced. From there one substitutes $\omega$ (the cube root of unity, $\omega^2 + \omega + 1 = 0$) or $i$ (so $i^2 + 1 = 0$) into $P$, getting $P(\omega) = \omega + 2$ and $P(i) = 2i + 1$, then reconstructs $P$ from those four constraints on its four unknown coefficients.

CCSS standards used (min grade 8)

6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Computing the final answer $1^2 + 2^2 + 3^2 + 3^2 = 23$.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Letting $a, b, c$ stand for the unknown coefficients of the linear quotients, then substituting back to write $P(x)$.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Distributing $(x^2+x+1)(ax+b)$ and $(x^2+1)(ax+c)$, then collecting like terms.)
6.EE.A.4 Identify when two expressions are equivalent (Testing degree $2$: two forms of $P$ must agree on every coefficient — and they cannot.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Solving the system $a+b=c$, $b=1$, $b+2=c+1$ for $a, b, c$.)

⭐ This AMC 10 problem only needs Grade 8 systems of equations you already know — try the smallest possible degree first (it fails by a clean $3=2$ contradiction), then write the next degree using letters for the unknown coefficients and match like terms to set up three tiny equations. Solve to get $P(x) = x^3 + 2x^2 + 3x + 3$, and the sum of squares of coefficients is $1+4+9+9 = 23$.