AMC 10 · 2002 · #14

Grade 11 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
logarithm-propertiesexponentsprime-factorization pattern-recognition ↑ Prerequisites: logarithm-properties
📏 Medium solution 💡 2 insights
Problem
A function is defined on the positive integers by f(n) = log₂₀₀₂ n². Three of its values are added together to form N = f(11) + f(13) + f(14). Decide which of the listed comparisons — with 1 and with 2 — the number N satisfies.

Pick an answer.

(A)
N<1
(B)
N=1
(C)
1<N<2
(D)
N=2
(E)
N>2
How to solve
Strategy Look for a Pattern

Each of f(11), f(13), f(14) is separately an ugly irrational number, so evaluating them one at a time leads nowhere useful. Tool #7 (Identify Subproblems) reframes the job: instead of three logarithms, compute the one logarithm they add up to, using the fact that logs with a common base add by multiplying their inputs. That turns the question into an arithmetic one — what is 11 · 13 · 14? — and tool #5 (Look for a Pattern) is what makes a solver suspect the answer will be the base itself, since contest problems pick inputs like these on purpose. Tool #15 (Organize Information in More Ways) supplies the exponential-form second route in the review.

1STEP 1

Evaluate f at the three inputs

Substituting gives three logarithms sharing base 2002; each alone is irrational, so combine them.

N = log₂₀₀₂ 11² + log₂₀₀₂ 13² + log₂₀₀₂ 14²
2STEP 2

Add the logarithms into one

Same-base logarithms add by multiplying inputs, so N = log₂₀₀₂(11² · 13² · 14²).

N = log₂₀₀₂(11² · 13² · 14²)
3STEP 3

Regroup into a single square

Reordering the factors turns three squares into one: (11 · 13 · 14)².

11² · 13² · 14² = (11 · 13 · 14)²
4STEP 4

Multiply the three inputs

The arithmetic is the punchline: 11 · 13 · 14 = 2002, exactly the base.

11 · 13 · 14 = 143 · 14 = 2002 = 2 · 7 · 11 · 13
5STEP 5

Read off the logarithm

So N = log₂₀₀₂(2002²) = 2 exactly, choice (D).

N = log₂₀₀₂(2002²) = 2
Answer
N=2
A decimal check supports the result: f(11) ≈ 0.6309, f(13) ≈ 0.6748, f(14) ≈ 0.6943, and those add to 2.0000 to four decimal places. It is worth being clear about what this check can and cannot do — decimals that round to 2.0000 would look almost identical if the true value were 1.9999 (choice C) or 2.0001 (choice E), so the estimate confirms but never proves. What settles it is the exact identity 11 · 13 · 14 = 2002, which makes the argument in the steps airtight rather than approximate. The size is also sensible: each of 11, 13, 14 is roughly the cube root of 2002 (since 12³ = 1728 and 13³ = 2197), so each f(n) = log₂₀₀₂ n² should be near 2/3, and three of them near 2.
💡Key takeaway

Logarithms with the same base add by multiplying what is inside them — so when a problem hands you numbers like 11, 13, and 14 next to a base of 2002, multiply them and see what happens.

  • Evaluate f at the three inputs
  • Add the logarithms into one
  • Regroup into a single square
  • Multiply the three inputs
  • Read off the logarithm

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