Competition · AMC preparation · step 4 of 4

AMC 10 · 2023A · #21

Grade 6 number-theory
polynomial-rootspolynomial-factoringfunction-evaluation identify-subproblemsconvert-to-algebrawork-backwards ↑ Prerequisites: polynomial-factoringfunction-evaluation
📏 Medium solution 💡 3 insights
Problem
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 mn\frac{m}{n} in lowest terms and find m+n.

Pick an answer.

(A)
41
(B)
43
(C)
45
(D)
47
(E)
49

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

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(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.

1STEP 1

Turn the clues into values

Turn each clue into a fact about P: P(1)=1, P(0)=0, P(9)=0, P(4)=0 — three roots plus one value condition.

P(1)=1, P(0)=0, P(9)=0, P(4)=0
2STEP 2

Collect the known roots

Minimal degree with one non-integer root means adding exactly one unknown root a, so P has four roots: 0, 4, 9, a.

Roots of P: 0, 4, 9, a
3STEP 3

Write P in factored form

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)
4STEP 4

Use the leftover clue

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

24(1-a)=1
5STEP 5

Solve for a

Solve the linear equation for a. Divide both sides by 24 to get 1-a=124\frac{1}{24}, so a=1-124\frac{1}{24}=(24−1)24\frac{(24-1)}{24}=2324\frac{23}{24}.

a=1-1/24=23/24
6STEP 6

Add the numerator and denominator

23 is prime and shares no factor with 24, so 2324\frac{23}{24} is already reduced: m=23, n=24, m+n=47 → (D).

m+n=23+24=47 → (D)
Answer
47
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)(1-2324\frac{23}{24})=24·124\frac{1}{24}=1. Every clue fires. The non-integer root 2324\frac{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 mn\frac{m}{n} where m+n is in that range; only 2324\frac{23}{24} matches.
💡Key takeaway

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=2324\frac{23}{24} and m+n=47.

  • Turn the clues into values
  • Collect the known roots
  • Write P in factored form
  • Use the leftover clue
  • Solve for a
  • Add the numerator and denominator

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