AMC 10 · 2021 · #14

Grade 11 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
logarithm-propertiesexponentssequences-arithmetictriangular-numbers identify-subproblemspattern-recognition ↑ Prerequisites: logarithm-properties
📏 Medium solution 💡 2 insights
Problem
Two long sums are multiplied together. The first adds logarithms whose base and argument both carry moving exponents, running up to twenty; the second adds a similar family running up to one hundred. Find the exact value of the product of these two sums.

Pick an answer.

(A)
21
(B)
$100\log_5 3$
(C)
$200\log_3 5$
(D)
$2{,}200$
(E)
$21{,}000$
How to solve
Strategy Organize Information in More Ways

There are 120 logarithms here and no chance of computing them one at a time. But every one of them is built from only two numbers, 3 and 5, dressed up in different powers. So the whole problem is a rewriting job: find one identity that strips the exponent off the base and the exponent off the argument, and every term collapses into a plain multiple of a single logarithm. Once that happens, each sum is just a number times log₅ 3 or a number times log₃ 5, and the two logarithms are reciprocals of each other. Handling the two sums as separate subproblems and only multiplying at the very end keeps the bookkeeping clean.

1STEP 1

Strip both exponents

One identity strips both exponents.

log_a^m bⁿ = (log bⁿ)/(log a^m) = (nlog b)/(mlog a) = n/m log_a b
2STEP 2

Tidy the first sum

The first becomes a sum of integers.

Σ_k=1²⁰ log₅^k 3^k² = Σ_k=1²⁰ k²/klog₅ 3 = (Σ_k=1²⁰ k)log₅ 3
3STEP 3

Evaluate the first sum

Add them by pairing.

Σ_k=1²⁰ k = (20 · 21)/2 = 210 → Σ_k=1²⁰ log₅^k 3^k² = 210log₅ 3
4STEP 4

Tidy the second sum

In the second, every term is identical.

Σ_k=1¹⁰⁰ log₉^k 25^k = Σ_k=1¹⁰⁰ k/klog₉ 25 = 100log₃² 5² = 100 · 2/2log₃ 5 = 100log₃ 5
5STEP 5

Multiply and let the logs cancel

Multiplying gives 21,000.

log₅ 3 · log₃ 5 = (log 3)/(log 5)·(log 5)/(log 3) = 1 → (210log₅ 3)(100log₃ 5) = 21000
Answer
21,000
Check the size with decimals. Since log₅ 3 ≈ 0.683, the first sum is about 210 × 0.683 ≈ 143. Since log₃ 5 ≈ 1.465, the second sum is about 100 × 1.465 ≈ 147. Their product is about 143 × 147 ≈ 21000, matching the exact answer. The choices also separate cleanly by size: 21 and 2200 are far too small, while 100log₅ 3 ≈ 68 and 200log₃ 5 ≈ 293 are smaller still, so only 21,000 is even in the right range. The clean cancellation also explains why the answer is exactly 20 · 21/2 × 100 = 21000 with no leftover logarithm.
💡Key takeaway

A power on the argument climbs on top and a power on the base drops underneath, so once every logarithm is written over the same two numbers, the ugly parts cancel and only counting is left.

  • Strip both exponents with one identity
  • Turn the first sum into a sum of integers
  • Add 1 through 20 by pairing
  • See that the second sum has constant terms
  • Multiply and let the logarithms cancel

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