AMC 10 · 2018 · #25
Grade 8 algebraPick an answer.
AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Move the floor term aside
Subtracting the floor term gives x² = 10000/x, where {x} = x - ⌊ x ⌋ is the fractional part in [0,1).
Whatever is left after you strip off the whole-number part of x is the fractional part, and that is all the floor term is really measuring.
6.EE.A.3Convert To AlgebraBound how big x can be
Since {x} is in [0,1), the right side stays under 10000, forcing x² ＜ 10000, so every solution is trapped in -100 ＜ x ＜ 100.
A square below 10000 means the number itself is squeezed between -100 and 100.
8.EE.A.2Use Matrix LogicZoom into one interval
On [n, n+1) the floor is n, so the equation is x² = 10000(x - n); the slow parabola meets the steep line at most once here.
Over one unit step the steep line either catches up to the slow-moving parabola exactly once, or never.
8.F.A.1Draw A DiagramFind which intervals give a crossing
The line overtakes the parabola inside the interval exactly when (n+1)² ＜ 10000; otherwise the parabola stays on top, no solution.
A solution exists precisely when the steep line out-climbs the parabola before the interval runs out.
6.EE.B.5Solve An Easier Related ProblemCount the good intervals
Solving (n+1)² ＜ 10000 gives n = -100, …, 98, that is 199 integers, each yielding one real x — so 199 solutions, choice (C).
One solution per qualifying integer n, so counting the integers counts the solutions.
6.EE.B.8Use Matrix LogicSplit 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