AMC 10 · 2011 · #20

Grade 8 algebranumber-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

quadratic-equationssystems-of-equationsdivisibility-rules pattern-recognitionbound-inequality-then-enumerate ↑ Prerequisites: systems-of-equations

📏 Long solution 💡 3 insights

Problem

A whole-coefficient quadratic vanishes at one input and lands inside narrow windows at two others. Locate a far-out value.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Look for a Pattern

Tool #15 (Organize Information in More Ways): f(1)=0 says a+b+c=0, so c=-a-b; substituting that back turns every value of f into an expression in a and b only. Tool #5 (Look for a Pattern): once c is gone, f(7)=6(8a+b), f(8)=7(9a+b), and f(100)=99(101a+b) — every f(n) carries the factor n-1. That single pattern is the engine of the whole problem: it makes f(7) a multiple of 6 and f(8) a multiple of 7, and it hands over f(100) at the end. Tool #3 (Eliminate Possibilities): a narrow window plus a divisibility requirement usually leaves one survivor, and here each window leaves exactly one. Tool #7 (Identify Subproblems): the leftover work is the small subproblem of solving two linear equations in a and b. Tool #6 (Guess and Check): the elimination only proves that one candidate could work, so the candidate has to be plugged back into every condition before it can be trusted.

1STEP 1

Use f(1)=0 to remove c

The known root removes one coefficient.

f(1)=a+b+c=0 → c=-a-b → f(x)=ax²+bx-a-b

2STEP 2

Every f(n) is a multiple of n-1

Every value then carries a visible factor.

f(7)=6(8a+b), f(8)=7(9a+b), f(100)=99(101a+b)

3STEP 3

The window for f(7) leaves one value

The first window holds exactly one multiple.

6 ∣ f(7), 50 ＜ f(7) ＜ 60 → f(7)=54 → 8a+b=9

4STEP 4

The window for f(8) leaves one value

The second window holds exactly one too.

7 ∣ f(8), 70 ＜ f(8) ＜ 80 → f(8)=77 → 9a+b=11

5STEP 5

Two equations pin a and b

Two equations pin both remaining coefficients.

8a+b=9 ; 9a+b=11 → a=2, b=-7, c=5

6STEP 6

Check the candidate really works

The candidate satisfies every condition.

f(1)=0, f(7)=54∈(50,60), f(8)=77∈(70,80)

7STEP 7

Compute f(100) and place it

The far value places it at 3, choice (C).

f(100)=99 · 195=19305, 5000 · 3 ＜ 19305 ＜ 5000 · 4 → k=3 (C)

Answer

Size check first: for large x the 2x² term dominates, so f(100) should be a bit under 2 · 100²=20000, and the -7x term pulls it down by 700. That predicts roughly 19300, matching 19305 exactly. Because the value lands just below 20000 rather than above it, k=3 and not 4 — the near miss is the whole difference between (C) and (D), which is why the exact arithmetic matters and an estimate alone would be unsafe. The smaller choices fail too: k=1 or k=2 would need f(100) ＜ 15000, forcing a ≤ 1, but a=1 with 8a+b=9 gives b=1 and then 9a+b=10 ≠ 11. Choice (E) would need f(100) ＞ 25000, roughly a ≥ 3, which breaks both windows. A second internal check: the second difference of a quadratic is 2a, and f(6)=35, f(7)=54, f(8)=77 give first differences 19 and 23, so the second difference is 4=2 · 2, confirming a=2.

💡Key takeaway

Because f(1)=0, every value f(n) is a multiple of n-1 — so the short windows for f(7) and f(8) each trap exactly one legal number, and those two numbers pin the whole parabola.

Use f(1)=0 to remove c
Every f(n) is a multiple of n-1
The window for f(7) leaves one value
The window for f(8) leaves one value
Two equations pin a and b
Check the candidate really works
Compute f(100) and place it