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2004 AMC 12A Problem 17

Problem 17 of 25IntermediateAlgebra

Let ff be a function with the following properties: (i) f(1)=1,f(1) = 1, and (ii) f(2n)=nf(n)f(2n) = n \cdot f(n) for any positive integer n.n. What is the value of f(2100)?f(2^{100})?

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Solution

Applying f(2n)=nf(n)f(2n) = n \cdot f(n) with n=2k,n = 2^{k}, we get f(2k+1)=2kf(2k).f(2^{k+1}) = 2^{k} \cdot f(2^{k}). Unwinding from f(21)=f(2)=1f(1)=20,f(2^1) = f(2) = 1 \cdot f(1) = 2^0, the exponents accumulate: f(2n)=20+1+2++(n1)=2n(n1)2. \begin{aligned} f(2^n) &= 2^{0 + 1 + 2 + \cdots + (n-1)} \\ &= 2^{\frac{n(n-1)}{2}}. \end{aligned} Therefore f(2100)=2100992=24950.f(2^{100}) = 2^{\frac{100 \cdot 99}{2}} = 2^{4950}. Thus, the correct answer is D.

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Concepts: functional equation · recursion · summation

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