AMC 10 · 2018 · #24

Grade 8 algebra
Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area
floor-functionquadratic-equationssign-analysis convert-to-algebracaseworkbound-inequality-then-enumerate ↑ Prerequisites: floor-function
📏 Medium solution 💡 3 insights
Problem
The floor of a real number is the greatest integer not larger than it. Count how many real numbers satisfy the equation x squared plus ten thousand times the floor of x equals ten thousand times x.

Pick an answer.

(A)
197
(B)
198
(C)
199
(D)
200
(E)
201
How to solve
Strategy Convert to Algebra

Tool #13 (Convert to Algebra): the messy part is the floor symbol, so the first move is to move it to one side and recognise x - ⌊ x ⌋ as the fractional part. That single rewrite turns the equation into x² = 10000{x}, which is far easier to read. Tool #4 (Introduce a Variable): name the integer part n = ⌊ x ⌋ so the whole real line breaks into separate intervals [n, n+1), one for each integer n. Tool #9 (Solve an Easier Related Problem): instead of the whole line at once, ask the small question 'how many solutions live in one interval [n, n+1)?' and then count the intervals that work. Tool #1 (Draw a Diagram): picturing the parabola y = x² against the rising sawtooth y = 10000{x} shows at a glance that each good interval gives exactly one crossing.

1STEP 1

Move the floor term aside

Moving the floor reveals the fractional part.

x² = 10000 (x - ⌊ x ⌋) = 10000{x}
2STEP 2

Bound how big x can be

The fractional part stays below one, bounding x.

0 ≤ {x} ＜ 1 → 0 ≤ x² ＜ 10000 → -100 ＜ x ＜ 100
3STEP 3

Zoom into one interval

Zoom into one integer interval.

x² = 10000(x-n), x ∈ [n, n+1)
4STEP 4

Find which intervals give a crossing

Decide which intervals give a crossing.

exactly one solution in [n, n+1) ⇔ (n+1)² ＜ 10000
5STEP 5

Count the good intervals

Counting them gives 199.

(n+1)² ＜ 10000 ⇔ n ∈ {-100, …, 98} → 199 solutions → (C)
Answer
199
The band -100 ＜ x ＜ 100 spans 200 unit intervals (integer parts n = -100 through 99). All but one of them produce a solution, and the lost one is n = 99, where (99+1)² = 10000 is not strictly less than 10000, so the steep line never quite passes the parabola. Dropping that single interval from 200 leaves 199, which lands on (C). The trap answers 197, 198, 200, 201 are exactly the kinds of numbers you get from an off-by-one slip in the endpoints, so the careful boundary check is what pins the count.
💡Key takeaway

Split a floor-function equation into one unit interval at a time, check each interval for a single crossing, then count the intervals that work.

  • Move the floor term aside
  • Bound how big x can be
  • Zoom into one interval
  • Find which intervals give a crossing
  • Count the good intervals

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