AMC 10 · 2014 · #15
Grade 11 number-theoryPick an answer.
The expression looks analytic — logarithms, the constant e — but the question asked is pure number theory. So the plan has two focus shifts. First, Tool #15 (Organize Information in More Ways) rewrites the sum of logarithms as a single logarithm of a product, and Tool #11 (Work Backwards) undoes the ln with e; after that no logarithms remain and e^p is an explicit product. Second, Tool #16 (Change Focus): never compute that product. The question only asks how many factors of 2 it holds, so track the exponent of 2 and ignore everything else. Tool #7 (Identify Subproblems) makes that count easy, because the exponent of 2 in a product is the sum of the exponents in the factors, so each k^k can be handled on its own. Finally, "largest" is a maximality claim, not just a divisibility claim, so Tool #14 (Extreme Principle) closes the argument by showing what is left after pulling out 2¹⁶ is odd — which is the part a quick count skips.
Pull each coefficient inside the log
Each coefficient rises into an exponent.
Multiplying a logarithm by k means multiplying the original number by itself k times, so the coefficient can move up into the exponent.
Multiplying a logarithm by a whole number means multiplying the original number by itself that many times.
▸ Why?
A logarithm counts how many times the base is used, so multiplying that count repeats the whole number.
▸ Why?
Repeating the same factor a fixed number of times is a count of equal groups, which multiplying records.
Collapse the sum, then undo the log
The sum collapses into one product.
Logarithms turn multiplication into addition, so reading that backwards turns the sum right back into one big product.
11.F-LE.A.4Work BackwardsTrack only the twos
Counting twos is additive across a product.
Factors of 2 never appear or disappear when you multiply — they just pile up, so counting them is pure addition.
8.EE.A.1Change Focus Count The ComplementCount the twos base by base
Only three bases contribute any twos.
Odd bases can never produce a factor of 2, so all the twos come from 2, 4, and 6 — and 4 pays double each time.
4.OA.B.4Identify SubproblemsProve no seventeenth two exists
What remains is odd, so the answer is 2 to the 16.
Once the leftover factor is odd, the supply of twos is genuinely exhausted — there is nothing left to halve.
4.OA.B.4Extreme PrincipleA sum of logarithms is a product wearing a disguise; strip the disguise, count how many 2s each factor donates, and then check that what is left over is odd so you know you found every last one.
- Pull each coefficient inside the log
- Collapse the sum, then undo the log
- Track only the twos
- Count the twos base by base
- Prove no seventeenth two exists