AMC 10 · 2020 · #17

Grade 8 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

polynomial-rootssign-analysissystematic-enumerationperfect-squaressequences-arithmetic easier-related-problempattern-recognitioncasework ↑ Prerequisites: polynomial-rootssign-analysis

📏 Long solution 💡 3 insights

Problem

The polynomial P(x) = (x-1)(x-4)(x-9)…(x-10000) has roots at the 100 perfect squares 1, 4, 9, …, 100² = 10000. Count how many integers n make P(n) ≤ 0 — that is, either zero or negative.

Pick an answer.

(A)

4900

(B)

4950

(C)

5000

(D)

5050

(E)

5100

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem): 100 factors are too many to track at once — replace 100 with a tiny number like 4 and look at Q(x) = (x-1)(x-4)(x-9)(x-16). Map out which integers make Q(n) ≤ 0. Tool #5 (Pattern): the signs of P(n) alternate +,-,+,-,… between consecutive square-roots, so the negative-or-zero intervals are exactly [1², 2²], [3², 4²], …, [99², 100²] — 50 intervals. Tool #7 (Subproblems): count integers in one general interval [(2k-1)², (2k)²], then sum over k = 1, …, 50.

1STEP 1

Shrink 100 factors to a 4-root toy Q: as n passes each root one factor changes sign, so Q's sign flips at every root.

Q(x) = (x-1)(x-4)(x-9)(x-16)

2STEP 2

The sign of P alternates, so P(n) ≤ 0 exactly on the closed intervals [1²,2²], [3²,4²], …, [99²,100²] — 50 intervals in all.

[(2k-1)², (2k)²] for k = 1, 2, …, 50

3STEP 3

In a general interval [(2k-1)²,(2k)²], the integer count is (2k)²-(2k-1)²+1 = (4k-1)+1 = 4k.

(2k)² - (2k-1)² + 1 = (4k-1) + 1 = 4k

4STEP 4

Sanity check: [1,4] holds 1,2,3,4 = 4·1, and [9,16] holds 8 integers = 4·2 — the 4k formula matches.

k=1: 4, k=2: 8, k=3: 12

5STEP 5

Sum over k=1…50: Σ4k = 4· $\frac{50·51}{2}$ = 4·1275 = 5100, choice (E).

Σ_k=1⁵⁰ 4k = 4 ·

\frac{50 · 51}{2}

= 5100 → (E)

Answer

5100

Compare to the largest root: 100² = 10,000. The total integers from 1 to 10,000 is 10,000, and the bad zones are roughly half of them — about 5,000. Our answer 5100 is right at that ballpark, and the slight excess comes from including the endpoint perfect squares (where P(n) = 0). The other choices (4900, 4950, 5000, 5050) sit one or two perfect-square-endpoints away, which is why the problem is built around this exact count.

💡Key takeaway

This AMC 10 problem only needs Grade 8 sign-pattern reasoning you already know — every time n crosses a perfect square the sign flips, so the bad zones [1², 2²], [3², 4²], …, [99², 100²] hold 4 + 8 + 12 + … + 200 = 5100 integers. The answer is (E).

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