AMC 10 · 2023 · #21

Grade 6 number-theory

polynomial-rootspolynomial-factoringfunction-evaluation identify-subproblemsconvert-to-algebrawork-backwards ↑ Prerequisites: polynomial-factoringfunction-evaluation

📏 Medium solution 💡 3 insights

Problem

Let $P(x)$ be the unique polynomial of minimal degree with the following properties:

$P(x)$ has a leading coefficient $1$ ,
$1$ is a root of $P(x)-1$ ,
$2$ is a root of $P(x-2)$ ,
$3$ is a root of $P(3x)$ , and
$4$ is a root of $4P(x)$ .

The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. What is $m+n$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: There is a polynomial $P(x)$ with leading coefficient $1$ and the smallest possible degree such that: $1$ is a root of $P(x)-1$, $2$ is a root of $P(x-2)$, $3$ is a root of $P(3x)$, and $4$ is a root of $4P(x)$. All roots of $P(x)$ are integers except one. Write that non-integer root as $\frac{m}{n}$ in lowest terms and find $m+n$.

Givens: $P(x)$ has leading coefficient $1$ and minimal possible degree; Condition 1: $P(1)-1=0$, i.e. $P(1)=1$; Condition 2: $P(2-2)=0$, i.e. $P(0)=0$ (so $0$ is a root of $P$); Condition 3: $P(3\cdot 3)=0$, i.e. $P(9)=0$ (so $9$ is a root of $P$); Condition 4: $4P(4)=0$, i.e. $P(4)=0$ (so $4$ is a root of $P$); All roots except one are integers; Answer choices: (A) $41$, (B) $43$, (C) $45$, (D) $47$, (E) $49$

Unknowns: The non-integer root $\frac{m}{n}$ of $P(x)$ in lowest terms, and the value $m+n$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #11 Work Backwards, #13 Convert to Algebra

Each of the four bullets is a tiny puzzle of the form "$k$ is a root of [something built from $P$]" — Tool #7 (Identify Subproblems) reads each clue separately and turns it into a single fact about $P$. Three of the four facts say $P(\text{something})=0$, which gives integer roots; the fourth says $P(1)=1$, a value condition. Tool #11 (Work Backwards) is the right move for that last clue: we already know three roots, so we write $P$ from the back (factored form), substitute $x=1$, and undo the arithmetic to recover the missing root. Tool #13 (Convert to Algebra) only needs to solve a single linear equation at the very end.

Execute — Answer: D

#7 Identify Subproblems 6.EE.B.5 Step 1

Translate each clue into one fact about $P$.
"$1$ is a root of $P(x)-1$" means $P(1)-1=0$, so $P(1)=1$.
"$2$ is a root of $P(x-2)$" means $P(2-2)=P(0)=0$.
"$3$ is a root of $P(3x)$" means $P(3\cdot 3)=P(9)=0$.
"$4$ is a root of $4P(x)$" means $4P(4)=0$, so $P(4)=0$.

$$P(1)=1,\quad P(0)=0,\quad P(9)=0,\quad P(4)=0$$

💡 Grade 6 "a root is just a value that makes the expression $0$" — read each clue, plug in, write what it says.

#7 Identify Subproblems 6.EE.B.5 Step 2

Collect the known roots.
The conditions force $0, 4, 9$ all to be roots of $P$, but they do not yet account for the requirement $P(1)=1$.
Because $P$ has minimal degree and one non-integer root, we add a single unknown root $a$.
So $P$ has exactly four roots: $0, 4, 9, a$.

$$\text{Roots of } P: \;0,\;4,\;9,\;a$$

💡 Three roots are nailed down; we add the fewest extra roots we can — exactly one more — and call it $a$.

#13 Convert to Algebra 6.EE.A.2 Step 3

Write $P$ in factored form.
With leading coefficient $1$ and roots $0,4,9,a$, the factor theorem gives $P(x)=(x-0)(x-4)(x-9)(x-a)=x(x-4)(x-9)(x-a)$.

$$P(x)=x(x-4)(x-9)(x-a)$$

💡 Grade 6 "write an expression that records the calculation" — each root contributes one factor.

#11 Work Backwards 6.EE.B.7 Step 4

Use the leftover clue $P(1)=1$ to pin down $a$.
Plug in $x=1$: $P(1)=1\cdot(1-4)\cdot(1-9)\cdot(1-a)=(-3)(-8)(1-a)=24(1-a)$.
Set equal to $1$.

$$24(1-a)=1$$

💡 Work backwards through the factored form: undo the multiplications around $a$ until $a$ is alone.

#13 Convert to Algebra 6.EE.B.7 Step 5

Solve the linear equation for $a$.
Divide both sides by $24$ to get $1-a=\frac{1}{24}$, so $a=1-\frac{1}{24}=\frac{24-1}{24}=\frac{23}{24}$.

$$a=1-\frac{1}{24}=\frac{23}{24}$$

💡 One-step linear equation — divide, then subtract — Grade 6 stuff.

#7 Identify Subproblems 4.OA.B.4 Step 6

Confirm $\frac{23}{24}$ is in lowest terms and finish.
$23$ is prime and does not divide $24$, so $\gcd(23,24)=1$.
Therefore $m=23, n=24$, giving $m+n=47$, which is choice (D).

$$m+n=23+24=47\;\Rightarrow\;\textbf{(D)}$$

💡 Grade 4 "is this number prime?" — $23$ has no factors in common with $24$, so the fraction is already reduced.

[1] #7 6.EE.B.5 Translate each clue into one fact about $P$. "$1$ is a root of $P(x)-1$" means $

[2] #7 6.EE.B.5 Collect the known roots. The conditions force $0, 4, 9$ all to be roots of $P$,

[3] #13 6.EE.A.2 Write $P$ in factored form. With leading coefficient $1$ and roots $0,4,9,a$, th

[4] #11 6.EE.B.7 Use the leftover clue $P(1)=1$ to pin down $a$. Plug in $x=1$: $P(1)=1\cdot(1-4)

[5] #13 6.EE.B.7 Solve the linear equation for $a$. Divide both sides by $24$ to get $1-a=\frac{1

[6] #7 4.OA.B.4 Confirm $\frac{23}{24}$ is in lowest terms and finish. $23$ is prime and does no

Review

Reasonableness: Spot-check the roots. $P(0)=0$, $P(4)=0$, $P(9)=0$ are visible from the factored form. For the value condition: $P(1)=(1)(-3)(-8)(1-\tfrac{23}{24})=24\cdot\tfrac{1}{24}=1$. Every clue fires. The non-integer root $\tfrac{23}{24}$ sits between $0$ and $1$, very close to $1$, which is consistent with the equation $24(1-a)=1$ (a tiny remainder forces $a$ close to $1$). The choices $41,43,45,47,49$ correspond to fractions $m/n$ where $m+n$ is in that range; only $\tfrac{23}{24}$ matches.

Alternative: Tool #6 (Guess and Check) on the answer choices: each choice $m+n$ paired with a likely non-integer root $\frac{m}{n}$ near $1$ can be plugged into $24(1-a)=1$ to see which works. $a=\frac{23}{24}$ gives $24\cdot\frac{1}{24}=1$ on the nose; the other candidate fractions miss.

CCSS standards used (min grade 6)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Confirming $\gcd(23,24)=1$ so that $\tfrac{23}{24}$ is already in lowest terms.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Building the factored expression $P(x)=x(x-4)(x-9)(x-a)$ from the roots.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values that make the statement true (Translating each "$k$ is a root" clue into a $P(\text{value})=0$ or $P(\text{value})=1$ statement and listing the forced roots.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $px = q$ (Solving the one-step equation $24(1-a)=1$ for the unknown root $a$.)

⭐ This AMC 10 puzzle melts once you turn each clue into one fact about $P$ — three of them say "this number is a root" and the last says "$P(1)=1$" — then it is just a Grade 6 one-step equation $24(1-a)=1$ that gives $a=\tfrac{23}{24}$ and $m+n=47$.