AMC 10 · 2020 · #17

Grade 8 arithmetic

Archetype Small-Domain Constrained Search

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

📏 Long solution 💡 3 insights

Problem

Define $P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).$ How many integers $n$ are there such that $P(n)\leq 0$ ?

$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$

Pick an answer.

(A)

4900

(B)

4950

(C)

5000

(D)

5050

(E)

5100

View mode:

Toolkit + CCSS Solution

Understand

Restated: The polynomial $P(x) = (x-1)(x-4)(x-9)\cdots(x-10000)$ has roots at the 100 perfect squares $1, 4, 9, \ldots, 100^2 = 10000$. Count how many integers $n$ make $P(n) \le 0$ — that is, either zero or negative.

Givens: $P(x)$ is a product of 100 factors $(x - k^2)$ for $k = 1, 2, \ldots, 100$; The 100 distinct roots are the perfect squares $1, 4, 9, 16, \ldots, 10000$; Choices: (A) $4900$, (B) $4950$, (C) $5000$, (D) $5050$, (E) $5100$

Unknowns: Number of integers $n$ satisfying $P(n) \le 0$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #5 Look for a Pattern, #7 Identify Subproblems

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 $Q(n) \le 0$. Tool #5 (Pattern): the signs of $P(n)$ alternate $+,-,+,-,\ldots$ between consecutive square-roots, so the negative-or-zero intervals are exactly $[1^2, 2^2], [3^2, 4^2], \ldots, [99^2, 100^2]$ — 50 intervals. Tool #7 (Subproblems): count integers in one general interval $[(2k-1)^2, (2k)^2]$, then sum over $k = 1, \ldots, 50$.

Execute — Answer: E

#9 Solve an Easier Related Problem 6.NS.C.7 Step 1

Shrink the problem from 100 factors to 4.
Let $Q(x) = (x-1)(x-4)(x-9)(x-16)$, with roots $1, 4, 9, 16$.
As $n$ increases past each root, exactly one factor flips from negative to positive, so the sign of $Q(n)$ flips each time.
For $n < 1$, all four factors are negative → product is positive (even number of negatives).
For $1 \le n \le 4$ one factor is non-negative and three are negative → product $\le 0$.
For $4 < n < 9$ two factors positive, two negative → positive.
The negative-or-zero zones are $[1, 4]$ and $[9, 16]$.

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

💡 Use a tiny version with 4 roots to see how signs flip.

#5 Look for a Pattern 8.F.B.5 Step 2

Spot the pattern: as $n$ crosses each successive perfect-square root, the sign of $P(n)$ flips.
With 100 factors (an even number), $P(n) > 0$ when $n$ is far below the smallest root.
So $P(n) \le 0$ exactly on the closed intervals $[1^2, 2^2], [3^2, 4^2], [5^2, 6^2], \ldots, [99^2, 100^2]$ — picking up between every odd-square and the next even-square.
There are $50$ such intervals.

$$[(2k-1)^2,\, (2k)^2] \text{ for } k = 1, 2, \ldots, 50$$

💡 Sign flips at every root — half the gaps between consecutive squares are "bad" zones.

#7 Identify Subproblems 7.EE.A.1 Step 3

Count integers in one general interval $[(2k-1)^2, (2k)^2]$.
The number of integers in a closed integer interval $[a, b]$ is $b - a + 1$.
So the count is $(2k)^2 - (2k-1)^2 + 1$.
Use the difference of squares: $(2k)^2 - (2k-1)^2 = (2k + 2k - 1)(2k - (2k-1)) = (4k-1)(1) = 4k - 1$.
Adding $1$: each interval contributes $4k$ integers.

$$(2k)^2 - (2k-1)^2 + 1 = (4k-1) + 1 = 4k$$

💡 The interval $[(2k-1)^2, (2k)^2]$ has length $4k - 1$, so $4k$ integer endpoints inside it.

#5 Look for a Pattern 5.OA.B.3 Step 4

Sanity-check on $k=1$: interval $[1, 4]$ has integers $1, 2, 3, 4$ — exactly $4 \cdot 1 = 4$.
On $k=2$: interval $[9, 16]$ has $9, 10, 11, 12, 13, 14, 15, 16$ — exactly $4 \cdot 2 = 8$.
The formula matches the small cases.

$$k=1: 4,\quad k=2: 8,\quad k=3: 12$$

💡 Quick check on tiny cases confirms the $4k$ count.

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

Sum the counts over $k = 1, 2, \ldots, 50$.
Total $= \sum_{k=1}^{50} 4k = 4 \cdot \sum_{k=1}^{50} k = 4 \cdot \tfrac{50 \cdot 51}{2} = 4 \cdot 1275 = 5100$.

$$\sum_{k=1}^{50} 4k = 4 \cdot \tfrac{50 \cdot 51}{2} = 5100 \;\Rightarrow\; \textbf{(E)}$$

💡 Add $4 + 8 + 12 + \cdots + 200$ using the triangle-number trick.

[1] #9 6.NS.C.7 Shrink the problem from 100 factors to 4. Let $Q(x) = (x-1)(x-4)(x-9)(x-16)$, wi

[2] #5 8.F.B.5 Spot the pattern: as $n$ crosses each successive perfect-square root, the sign o

[3] #7 7.EE.A.1 Count integers in one general interval $[(2k-1)^2, (2k)^2]$. The number of integ

[4] #5 5.OA.B.3 Sanity-check on $k=1$: interval $[1, 4]$ has integers $1, 2, 3, 4$ — exactly $4

[5] #7 6.EE.A.2 Sum the counts over $k = 1, 2, \ldots, 50$. Total $= \sum_{k=1}^{50} 4k = 4 \cdo

Review

Reasonableness: Compare to the largest root: $100^2 = 10{,}000$. 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.

Alternative: Tool #2 (Systematic List) plus Tool #14 (Finite Differences): list the counts $4, 8, 12, \ldots$ — this is an arithmetic sequence with first term $4$ and common difference $4$. With $50$ terms, the sum is $\tfrac{50}{2}(4 + 200) = 25 \cdot 204 = 5100$. Same answer, just framed as an arithmetic series.

CCSS standards used (min grade 8)

6.NS.C.7 Understand ordering and absolute value of rational numbers (Tracking how the sign of each factor $(n - k^2)$ flips as $n$ crosses each square-root.)
8.F.B.5 Describe qualitatively the functional relationship between two quantities (Recognizing the alternating sign pattern of $P(n)$ across consecutive square-root intervals.)
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Simplifying $(2k)^2 - (2k-1)^2 + 1$ via the difference-of-squares identity to get $4k$.)
5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships (Verifying the $4k$ rule against the small cases $k=1, 2, 3$.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Summing the arithmetic sequence $\sum_{k=1}^{50} 4k$ using the triangle-number formula.)

⭐ 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 $[1^2, 2^2], [3^2, 4^2], \ldots, [99^2, 100^2]$ hold $4 + 8 + 12 + \cdots + 200 = 5100$ integers. The answer is $\textbf{(E)}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 8)

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