AMC 10 · 2006 · #20

Grade 11 probability

Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Region

floor-functionlogarithm-propertiesgeometric-probabilitygeometric-series-infinite identify-subproblemspattern-recognition ↑ Prerequisites: floor-functionlogarithm-properties

📏 Long solution 💡 4 insights

Problem

A number is picked uniformly from between zero and one, and compared with four times itself. Find the probability that both have the same number of digits before the decimal shift.

Pick an answer.

(A)

$\frac 18$

(B)

$\frac 3{20}$

(C)

$\frac 16$

(D)

$\frac 15$

(E)

$\frac 14$

How to solve

Strategy Identify Subproblems

The interval (0,1) is awkward because the interesting behaviour piles up near 0: there is no single formula covering all of it at once. Tool #7 (Identify Subproblems) cuts it at the powers of ten into windows [10^k,10^k+1), which is the right cut because ⌊log₁₀⌋ is constant on exactly those windows and nowhere larger. Tool #4 (Introduce a Variable) names that constant k, turning an equation about floors into plain inequalities about x. Tool #16 (Change Focus) asks what can go wrong rather than what must go right: because 4 ＜ 10, the only failure is 4x spilling over the top edge of its own window. Tool #9 (Solve an Easier Related Problem) then settles one window completely, and Tool #5 (Look for a Pattern) catches the fact that makes the whole infinite pile-up harmless — the favorable fraction of a window is the same number for every k.

1STEP 1

Read the floor as a power-of-ten window

The floor names a window between powers of ten.

⌊log₁₀t⌋=k⇔ k ≤ log₁₀t ＜ k+1⇔ 10^k ≤ t ＜ 10^k+1

2STEP 2

Multiplying by 4 moves you at most one window

Multiplying by four moves you at most one window.

10^k ≤ x ＜ 10^k+1 → 10^k ＜ 4x ＜ 10^k+2; 0 ＜ log₁₀₄ ＜ 1 → ⌊log₁₀₄x⌋-⌊log₁₀x⌋∈{0,1}

3STEP 3

Settle a single decade exactly

Inside one window the good part is a clean interval.

favorable in decade k=[10^k, (10^k+1)/4)=[10^k, 2.5 · 10^k), length 1.5 · 10^k, decade length 9 · 10^k

4STEP 4

Every decade gives the same fraction

Every window gives the same fraction.

(1.5 · 10^k)/(9 · 10^k)=(10/4-1)/(10-1)=3/2/9=1/6 for every integer k

5STEP 5

The decades tile (0,1) with nothing left over

The windows tile the range with nothing left over.

(0,1)=bigcup_k ≤ -1[10^k,10^k+1) (disjoint), Σ_k ≤ -19 · 10^k=9·1/10/(1-1/10)=1

6STEP 6

Add the favorable lengths

Adding the good lengths gives 1/6.

Σ_k ≤ -1 3/2 · 10^k=3/2·1/10/(1-1/10)=3/2·1/9=1/6

7STEP 7

Turn length into probability

The total length is one, so the probability is 1/6, choice (C).

P=(favorable length)/(total length)=1/6/1=1/6 → (C)

Answer

1/6

Three checks agree. First, size: 1/6≈ 0.167 is a plausible slice, and sampling x on a fine grid across (0,1) makes the condition hold about 16.7% of the time. Second, the trap: choice (B) 3/20=0.15 is exactly the favorable length of the single decade [0.1,1), namely [0.1,0.25). Anyone who stops after the biggest window lands on 3/20, and the gap between 3/20 and 1/6 is precisely the contribution of all the smaller windows — so (B) is the answer to a different question. Third, the load-bearing assumption is visible: the answer is 1/6 because x is uniform inside each decade. Any other distribution that is uniform within each decade gives 1/6 too, no matter how it weights the decades, while a distribution uniform in log₁₀x instead would give 1-log₁₀₄≈ 0.398. Finally, the boundary points x=2.5 · 10^k, where 4x is exactly a power of ten, form a countable set of total length 0, so including or excluding them cannot change the answer.

💡Key takeaway

Multiplying by 4 can push a number across at most one power-of-ten line, and in every power-of-ten window the part that stays put is always the same one sixth, so the probability is 1/6.

Read the floor as a power-of-ten window
Multiplying by 4 moves you at most one window
Settle a single decade exactly
Every decade gives the same fraction
The decades tile (0,1) with nothing left over
Add the favorable lengths
Turn length into probability