AMC 10 · 2002 · #22

Grade 11 algebra
Archetype Telescoping Collapse └ Sub-archetype Shifted Series Difference
logarithm-propertiesexponentsprime-factorization identify-subproblems ↑ Prerequisites: logarithm-properties
📏 Long solution 💡 3 insights
Problem
A rule builds one number for every integer n bigger than 1: take log_n 2002 and flip it over, so a_n = 1/(log_n 2002). Add four of these to get b = a₂ + a₃ + a₄ + a₅, and add five of them to get c = a₁₀ + a₁₁ + a₁₂ + a₁₃ + a₁₄. Find b - c.

Pick an answer.

(A)
-2
(B)
-1
(C)
$\frac{1}{2002}$
(D)
$\frac{1}{1001}$
(E)
$\frac 12$
How to solve
Strategy Organize Information in More Ways

Nine different logarithm bases with nothing in common looks hopeless, so the move is tool #15 (Organize Information in More Ways): rewrite every term so it sits in one fixed base. The reciprocal 1/(log_n 2002) is exactly log₂₀₀₂ n, and once all nine terms share the base 2002 the sums collapse into single logarithms. Tool #4 (Introduce a Variable) proves that flip honestly instead of quoting it: name the exponent and read the equation backwards. Tool #7 (Identify Subproblems) then splits the work — turn b into one logarithm, turn c into one logarithm, and only at the end compare them.

1STEP 1

Flip the reciprocal

Flipping the reciprocal swaps base and argument, so a_n = log₂₀₀₂ n and all terms share a base.

x = log_n 2002 ⇔ n^x = 2002 ⇔ n = 2002¹/x ⇔ log₂₀₀₂ n = 1/x = a_n
2STEP 2

Fold b into one logarithm

Same-base logarithms add by multiplying, so b = log₂₀₀₂(2 · 3 · 4 · 5).

b = log₂₀₀₂(2 · 3 · 4 · 5)
3STEP 3

Fold c into one logarithm

The same fold gives c = log₂₀₀₂(10 · 11 · 12 · 13 · 14).

c = log₂₀₀₂(10 · 11 · 12 · 13 · 14)
4STEP 4

Subtract, then cancel

Subtracting divides; cancelling factor by factor leaves 1/(11 · 13 · 14) inside.

b - c = log₂₀₀₂((2 · 3 · 4 · 5)/(10 · 11 · 12 · 13 · 14)) = log₂₀₀₂(1/(11 · 13 · 14))
5STEP 5

Recognize 2002

But that product is 2002, the base itself, so b - c = -1, choice (B).

11 · 13 · 14 = 2002 → b - c = log₂₀₀₂(2002⁻¹) = -1
Answer
-1
First, the sign is right: c has more terms than b and every one of them is larger, so b - c must be negative. Second, here is a check that never uses a single logarithm rule. Undo the logs into exponent form: 2002^b = 2 · 3 · 4 · 5 = 120 and 2002^c = 10 · 11 · 12 · 13 · 14 = 240240. Since 240240 = 120 · 2002, we get 2002^c = 2002^b · 2002¹ = 2002^b+1. The function t ↦ 2002^t takes each value only once, so c = b + 1, hence b - c = -1. Third, a numerical check with common logs: log 2002 ≈ 3.3016, so b ≈ 2.0792/3.3016 ≈ 0.6297 and c ≈ 5.3806/3.3016 ≈ 1.6297, giving b - c ≈ -1.0000. All three agree on (B).
💡Key takeaway

A flipped logarithm just swaps its two numbers: 1/(log_n N) = log_N n — so nine different bases become one base, and the sums melt into a single product.

  • Flip the reciprocal
  • Fold b into one logarithm
  • Fold c into one logarithm
  • Subtract, then cancel
  • Recognize 2002

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