AMC 10 · 2014 · #21

Grade 11 algebra

Archetype Telescoping Collapse └ Sub-archetype Product Ratio Cancellation

floor-functionlogarithm-propertiestelescoping-suminterval-arithmetic convert-to-algebraidentify-subproblemspattern-recognition ↑ Prerequisites: floor-functionlogarithm-properties

📏 Medium solution 💡 4 insights

Problem

A floor function freezes on each unit block, and the condition holds on part of each. Add the lengths.

Pick an answer.

(A)

(B)

$\dfrac{\log 2015}{\log 2014}$

(C)

$\dfrac{\log 2014}{\log 2013}$

(D)

$\dfrac{2014}{2013}$

(E)

$2014^{\frac1{2014}}$

How to solve

Strategy Introduce a Variable

The floor symbol is the only thing making f hard, and a floor is constant between consecutive integers. So Tool #4 (Introduce a Variable) names the two halves of x separately: k=⌊ x⌋ and the fractional part t=x-k. On the block k ≤ x ＜ k+1 the number k is frozen, and f turns into the ordinary exponential k(2014^t-1). Tool #13 (Convert to Algebra) then solves f(x) ≤ 1 for t with a logarithm. Tool #14 (Extreme Principle) is used deliberately on the two boundary questions this problem hides: whether the right endpoint is kept (the inequality is non-strict, so it is) and how long the kept piece can possibly be (short enough that it never reaches the next block, which is what makes the intervals disjoint and their lengths addable). Tool #7 (Identify Subproblems) settles which blocks exist, where the strict bound x ＜ 2014 matters. Tool #5 (Look for a Pattern) finishes the job: the lengths are logarithms of (k+1)/k, and that product telescopes.

1STEP 1

Freeze the floor on each block

On each block the floor is a constant.

x=k+t, k=⌊ x⌋inZ, 0 ≤ t ＜ 1 ⟹ f(x)=k(2014^t-1)

2STEP 2

Solve the inequality with a logarithm

A logarithm solves the inequality exactly.

k(2014^t-1) ≤ 1⇔ 2014^t ≤ (k+1)/k⇔ t ≤ log₂₀₁₄ (k+1)/k

3STEP 3

Check endpoints, then check disjointness

Each piece is short enough to stay inside its block.

f(k)=0, f (k+log₂₀₁₄ (k+1)/k)=1; length_k=log₂₀₁₄ (k+1)/k ≤ log₂₀₁₄₂≈ 0.0911 ＜ 1

4STEP 4

List the blocks that exist

The blocks run over a known range.

1 ≤ x ＜ 2014 ⟹ k∈{1,2,…,2013}

5STEP 5

Add the lengths; the product telescopes

The product collapses, leaving 1, choice (A).

Σ_k=1²⁰¹³log₂₀₁₄ (k+1)/k=log₂₀₁₄Π_k=1²⁰¹³(k+1)/k=log₂₀₁₄ 2014/1=1 ⟹ (A)

Answer

Size check first: each length log₂₀₁₄ (k+1)/k is positive and at most log₂₀₁₄₂≈ 0.0911, and the lengths shrink fast (at k=2013 the length is about 0.0000653), so 2013 of them summing to a modest number near 1 is believable, and adding them numerically gives 1.000000000000 to twelve digits. The two boundary decisions are what pin the value to exactly 1. If the right endpoint had been dropped, each interval would be half-open, which changes nothing about length, so the non-strict ≤ is safe. If instead the strict bound x ＜ 2014 were misread as x ≤ 2014, an extra block k=2014 of length log₂₀₁₄ 2015/2014 would join the sum and the total would become log₂₀₁₄₂₀₁₅=(log 2015)/(log 2014), which is choice (B). Choices (B), (C), (D), (E) are all just over 1, so they are precisely the near-misses of this problem; landing on the exact value 1 confirms the block list stopped in the right place. Spot check at k=1: the interval is [1, 1+log₂₀₁₄₂] and at its right end f(x)=1·(2-1)=1, satisfying f(x) ≤ 1 as claimed.

💡Key takeaway

Freeze the floor on each unit block: the part you keep is a short segment of length log₂₀₁₄ (k+1)/k starting at k, and since those fractions multiply out to 2014, all the pieces together measure exactly 1.

Freeze the floor on each block
Solve the inequality with a logarithm
Check endpoints, then check disjointness
List the blocks that exist
Add the lengths; the product telescopes

Freeze the floor on each block

Solve the inequality with a logarithm

Check endpoints, then check disjointness

List the blocks that exist

Add the lengths; the product telescopes

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