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)

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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 $1+4+9+9 = 23$.