AMC 10 · 2013 · #22

Grade 11 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Symmetric Function of Roots via Vieta

logarithm-propertiesvieta-formulasprime-factorizationoptimization convert-to-algebraidentify-subproblemsextremal-construction ↑ Prerequisites: logarithm-propertiesquadratic-equations

📏 Long solution 💡 4 insights

Problem

Two whole-number bases appear in one logarithmic equation whose root product must be whole. Find the bases' sum at the smallest product.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

272

How to solve

Strategy Introduce a Variable

The equation looks like it has two unknown logarithms, but log_n x and log_m x are the same number ln x read off two different scales. Naming L = ln x once (Introduce a Variable) collapses the equation into a quadratic in L, and then the product of the solutions is an exponentiated root-sum, which Vieta gives without ever solving the quadratic. That splits the problem cleanly in two (Identify Subproblems): part one is the algebra that produces the formula ⁸√(m⁷ n⁶); part two is a pure minimization over integers. The second part is where the real work is, and a minimum claim always has two halves (Extreme Principle): a lower bound nothing can beat, and one explicit pair that reaches it. Checking a small case and reading off a pattern is not a proof of minimality, so the bound is derived from prime exponents.

1STEP 1

Name the one moving part

One letter names the moving part.

L = ln x, a = ln m ＞ 0, b = ln n ＞ 0, log_m x = L/a, log_n x = L/b

2STEP 2

Clear the fractions

Clearing fractions gives a plain quadratic.

8L²/ab - 7L/b - 6L/a - 2013 = 0 → 8L² - (7a + 6b)L - 2013ab = 0

3STEP 3

Check the solutions exist

Its roots are real and distinct.

L₁ L₂ = -2013ab/8 ＜ 0 → (7a + 6b)² + 64 · 2013 ab ＞ 0 → L₁ ≠ L₂ both real, x_i = e^L_i ＞ 0

4STEP 4

Vieta, then undo the log

The sum of roots gives the wanted product.

L₁ + L₂ = (7a + 6b)/8 = (7ln m + 6ln n)/8 → P = x₁ x₂ = e^L₁ + L₂ = m⁷/8 n⁶/8 = ⁸√(m⁷ n⁶)

5STEP 5

Integrality is about exponents

Being whole is a condition on exponents.

P⁸ = m⁷ n⁶, m = Π p^α_p, n = Π p^β_p, P = Π p^γ_p → 8γ_p = 7α_p + 6β_p for every prime p

6STEP 6

A prime inside m costs four

Any prime in one base forces a floor.

α ≡ 2γ (mod 6), 7α ≤ 8γ, α ≥ 1 ⟹ γ ≥ 4 ⟹ P ≥ p⁴ ≥ 2⁴ = 16

7STEP 7

Reach the floor and add

Reaching that floor gives 12, choice (E).

γ = 4: α = 2, β = 3 → m = 4, n = 8, m⁷ n⁶ = 2¹⁴ · 2¹⁸ = 2³² = 16⁸ → P = 16, m + n = 12

Answer

Test (m,n) = (4,8) against the original equation without any of the machinery. Put t = log₂ x, so log₄ x = t/2 and log₈ x = t/3, and the equation becomes 8 · t/3 · t/2 - 7 · t/3 - 6 · t/2 - 2013 = 0, that is 4t² - 16t - 6039 = 0. Its roots sum to 4, so x₁ x₂ = 2⁴ = 16, matching the formula. Numerically t ≈ 40.91 and t ≈ -36.91, giving x₁ ≈ 2.06 × 10¹² and x₂ ≈ 7.76 × 10⁻¹², whose product is 16 — one huge solution and one tiny one, exactly the opposite-sign pattern step 3 predicted. A sweep of every pair with 2 ≤ m, n ≤ 3000 finds no integer product below 16, and finds 16 only at (4,8), which agrees with the proof that the exponents were forced. The distractors are instructive: (m,n) = (16,4) also makes the product an integer, namely 32, and gives m + n = 20, choice (B) — the trap for stopping at the first integer found instead of the smallest. Likewise (256,16) gives the integer 1024 and m + n = 272, choice (E). Answer (A) stands.

💡Key takeaway

Every logarithm in the equation is the same number wearing a different base, so name it once: the equation becomes a quadratic, and the sum of its roots is the logarithm of the product you were asked for.

Name the one moving part
Clear the fractions
Check the solutions exist
Vieta, then undo the log
Integrality is about exponents
A prime inside m costs four
Reach the floor and add