AMC 10 · 2008 · #23

Grade 11 number-theoryalgebra

Archetype Pair-Sum Invariant └ Sub-archetype Uniform Pair-Sum Recovery

logarithm-propertiesdivisor-countprime-factorizationpair-counting convert-to-algebraidentify-subproblems ↑ Prerequisites: logarithm-propertiesprime-factorization

📏 Medium solution 💡 3 insights

Problem

Every divisor of a power of ten has its logarithm taken, and all of them are added. The total is given. Find the exponent.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Organize Information in More Ways

The sum has no obvious first move because each individual logarithm is irrational. The fix is to stop viewing the divisors as a list of numbers and start viewing them as a square grid of exponent pairs (a,b), since every divisor of 2ⁿ 5ⁿ is 2^a 5^b. Once the sum runs over that grid, the logarithm rules convert it into a sum of exponents weighted by the two constants log₁₀ 2 and log₁₀ 5, which is just two copies of the same simple sum 0+1+…+n. The two constants then merge into 1 because 2 · 5 = 10, and every irrational piece disappears at once. What is left is a single equation in n, which is finished off by elimination over the integers rather than by testing the five answer choices.

1STEP 1

Name each divisor by its exponents

Every divisor is named by two exponents.

10ⁿ = 2ⁿ 5ⁿ, d ∣ 10ⁿ ⇔ d = 2^a 5^b with 0 ≤ a ≤ n, 0 ≤ b ≤ n

2STEP 2

Count the divisors

Counting them gives a perfect square.

τ(10ⁿ) = (n+1)(n+1) = (n+1)²

3STEP 3

Turn each logarithm into exponents

Each logarithm splits into a weighted sum.

log₁₀ d = log₁₀(2^a 5^b) = a log₁₀ 2 + b log₁₀ 5

4STEP 4

Add across the whole grid

Summing across the grid gives one expression.

S(n) = Σ_a=0ⁿ Σ_b=0ⁿ (a log₁₀ 2 + b log₁₀ 5) = (n(n+1)²)/2 log₁₀ 2 + (n(n+1)²)/2 log₁₀ 5

5STEP 5

Collapse the two constants into one

The two logarithms add to one and vanish.

S(n) = (n(n+1)²)/2(log₁₀ 2 + log₁₀ 5) = (n(n+1)²)/2 · 1 = (n(n+1)²)/2

6STEP 6

Solve for n without guessing

Factoring the total gives 11, choice (B).

(n(n+1)²)/2 = 792 ⇔ n(n+1)² = 1584 = 2⁴ · 3² · 11, 11 · 12² = 11 · 144 = 1584

Answer

Check n = 11 forward instead of backward. Then 10¹¹ has 12² = 144 divisors, and matching each divisor d with 10¹¹/d shows the logarithms are spread symmetrically about their midpoint 11/2, so the average logarithm is 5.5 and the total is 144 × 5.5 = 792, exactly as required. The formula also separates the five choices cleanly: S(11) = 792, S(12) = 1014, S(13) = 1274, S(14) = 1575, S(15) = 1920, so no other listed value comes close, and since n(n+1)² strictly increases for n ≥ 1, no unlisted n can produce 792 either. A rough size check agrees too: the 144 logarithms run from log₁₀ 1 = 0 up to log₁₀ 10¹¹ = 11, so the total has to land somewhere between 0 and 144 × 11 = 1584, and 792 is precisely the midpoint of that window, which is what perfect symmetry predicts.

💡Key takeaway

Divisors of 10ⁿ come in partners that multiply to 10ⁿ, so their logarithms average n/2, and the whole sum is just the number of divisors times n/2.

Name each divisor by its exponents
Count the divisors
Turn each logarithm into exponents
Add across the whole grid
Collapse the two constants into one
Solve for n without guessing