AMC 10 · 2025 · #24

Grade 11 algebra

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Periodic Placement Formula

logarithm-propertiesperiodic-functionsign-analysis bound-inequality-then-enumeratecasework ↑ Prerequisites: logarithm-properties

📏 Long solution 💡 5 insights

Problem

Count every real number x for which sin(20π x) and log₂₀(x) have the same value.

Pick an answer.

(A)

199

(B)

200

(C)

398

(D)

399

(E)

400

How to solve

Strategy Draw a Diagram

No algebra isolates x here, so I stop trying to solve and start counting: put both sides on one picture and count how often the fast wave crosses the slow log curve. The sine's ceiling and floor of ± 1 trap the whole search inside a finite window, which turns an infinite question into a finite one. Inside that window the wave repeats, so I count in repeat-units instead of one crossing at a time: each stretch between neighbouring zeros of the wave is a single hill or a single valley. Comparing signs then wipes out most of those units, and the survivors are all identical in shape, so each contributes the same small number of crossings. The last job is bookkeeping — checking whether any crossing sits on a seam and got charged to two units.

1STEP 1

Read it as two graphs

Turn it into graph intersections.

f(x) = sin(20π x), g(x) = log₂₀(x)

2STEP 2

Trap x inside a finite window

x is trapped between one twentieth and twenty.

-1 ≤ log₂₀(x) ≤ 1 ⇔ 1/20 ≤ x ≤ 20

3STEP 3

Cut the window into arches

Cut the window into arches.

T = 2π/20π = 1/10, arch width = 1/20, (20 - 1/20)/1/20 = 399

4STEP 4

Sign matching kills most arches

Only 200 arches have matching sign.

10 live valleys + 190 live hills = 200 live arches

5STEP 5

Each live arch is crossed twice

Each live arch is crossed twice.

200 × 2 = 400 counted crossings

6STEP 6

Remove the one double count

Removing the one double count leaves 399.

400 - 1 = 399

Answer

399

Each wrong choice is a real quantity from the count, and the count survives every one of those traps. The value 400 is the arch-by-arch total taken before the shared point at x = 1 is deduplicated; 398 would come from deducting that seam twice; 200 is the number of live arches rather than crossings; 199 is that same count off by one. The structural bookkeeping also closes: the window holds 399 arches, of which 200 are live and 199 are dead, and the split 10 + 190 matches the two half-counts. Graphing sin(20π x) - log₂₀(x) over [1/20, 20] and counting the places where it changes sign gives 399 as well. Finally, neither end of the window is a stray extra solution: at x = 1/20 the wave is 0 but the log is -1, and at x = 20 the wave is 0 but the log is 1.

💡Key takeaway

When a fast wave meets a slow curve, stop solving and start counting: fence in where they could possibly meet, chop the fence into single arches, throw out every arch on the wrong side of zero, give each survivor two crossings, then check whether any crossing landed on a seam and got counted twice.

Read it as two graphs
Trap x inside a finite window
Cut the window into arches
Sign matching kills most arches
Each live arch is crossed twice
Remove the one double count