AMC 10 · 2017 · #21

Grade 9 algebra
rational-root-theorempolynomial-rootsdivisibility-rulesfactors bound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: polynomial-rootsdivisibility-rules
📏 Medium solution 💡 3 insights
Problem
Whole-number roots of polynomials with coefficients from a set get added back to it. Count the final set.

Pick an answer.

(A)
4
(B)
5
(C)
7
(D)
9
(E)
11
How to solve
Strategy Eliminate Possibilities

The rule looks open-ended — the candidate roots range over all the integers, and there is no obvious reason the set should ever stop growing. So the question splits into two opposite halves, and both must be settled. The ceiling half is where Tool #3 (Eliminate Possibilities) does the heavy lifting: a single divisibility fact about polynomial roots shows that a newcomer must divide a number already present, which fences the whole process inside a short, explicit list of integers. That converts an infinite search into a finite one. The attainment half then runs Tool #11 (Work Backwards): for each number on that short list, start from the root you want and design a polynomial that has it, using only coefficients already in the set. Tool #2 (Make a Systematic List) keeps the two halves honest by recording the set after each addition, since a construction is legal only if its coefficients were present at the time it was used. When the ceiling and the constructions land on the same set, the count is proved, not guessed.

1STEP 1

A root must divide a coefficient

A whole root always divides a coefficient.

p(x) = x^k(a_k + a_k+1x + … + a_nxⁿ-k) → a_k = -x(a_k+1 + a_k+2x + … + a_nxⁿ-k-1) → x ∣ a_k
2STEP 2

The process is trapped inside the divisors of 10

That traps everything inside nine possible values.

divisors of 10: ± 1, ± 2, ± 5, ± 10; D = {-10,-5,-2,-1,0,1,2,5,10}, |D| = 9
3STEP 3

Build the two signs and -10

Simple polynomials build the first few.

10x + 10 = 0 → x = -1; (-x¹⁰ - x⁹ - … - x + 10)|_x=1 = -10 + 10 = 0; x + 10 = 0 → x = -10
4STEP 4

Reach 2, and the rest follows

Reaching one key value unlocks the rest.

2³ + 2 - 10 = 0; x + 2 = 0 → x = -2; 2x - 10 = 0 → x = 5; x + 5 = 0 → x = -5
5STEP 5

Check the order, not just the list

The building order also has to work.

{0,10} → -1 → 1 → -10 → 2 → -2 → 5 → -5
6STEP 6

Nothing more can be added

The final set has 9 members, choice (D).

S = {-10,-5,-2,-1,0,1,2,5,10}, |S| = 9 → (D)
Answer
9
The two halves meet, which is the real check: the divisibility argument says S can never contain more than the nine numbers {-10,-5,-2,-1,0,1,2,5,10}, and the explicit polynomials produce all nine, so the count is pinned from above and below at 9. Each construction survives substitution: 10(-1) + 10 = 0, the eleven coefficients -1,-1,…,-1,10 sum to 0 at x = 1, (-10) + 10 = 0, 2³ + 2 - 10 = 0, (-2) + 2 = 0, 2(5) - 10 = 0, and (-5) + 5 = 0. The wrong choices are exactly the places a solver stops too early: 4 after collecting {-1,0,1,10}, 5 after adding -10, 7 after adding 2 and -2 but missing that 2x - 10 unlocks 5. Choice 11 is impossible outright, since it would need a member outside the divisors of 10. One more sanity signal: apart from 0, the final set is symmetric under changing sign, which is forced once 1 is a member, because x + m always converts m into -m. That makes the size odd — one for 0 plus four ± pairs — matching 9.
💡Key takeaway

Find the ceiling first — a new number always has to divide a number already in the set, so nothing can escape the divisors of 10 — then build all nine of those numbers with real polynomials and stop.

  • A root must divide a coefficient
  • The process is trapped inside the divisors of 10
  • Build the two signs and -10
  • Reach 2, and the rest follows
  • Check the order, not just the list
  • Nothing more can be added