2004 AMC 10A Problem 24

Put $n = 2^{k-1}$ in rule (ii). Since $2n = 2^k$, the rule reads
$$f(2^k) = 2^{k-1} \cdot f(2^{k-1}) \qquad (k \ge 1).$$

Chain these from $k = 100$ down to $k = 1$:
$$f(2^{100}) = 2^{99} f(2^{99}) = 2^{99} \cdot 2^{98} f(2^{98}) = \cdots = 2^{99} \cdot 2^{98} \cdots 2^{1} \cdot 2^{0} \cdot f(1).$$
With $f(1) = 1$, the product is a single power of $2$ whose exponent is
$$99 + 98 + \cdots + 1 + 0 = \frac{99 \cdot 100}{2} = 4950.$$

The answer is $\boxed{\textbf{(D)}\ 2^{4950}}$.

A rule relating $f(2n)$ to $f(n)$ is a recursion along the powers of $2$; unwinding it multiplies together one factor per step, and multiplying powers of $2$ adds the exponents. Computing $f(2), f(4), f(8) = 1, 2, 8 = 2^{0}, 2^{1}, 2^{3}$ makes the triangular-number pattern $2^{0+1+\cdots+(k-1)}$ visible immediately.

Small cases: $f(2) = 1$, $f(4) = 2 \cdot 1 = 2$, $f(8) = 4 \cdot 2 = 8$, $f(16) = 8 \cdot 8 = 64 = 2^6$. The exponents $0, 1, 3, 6$ are triangular numbers $\binom{k}{2}$ for $k = 1, 2, 3, 4$, so $f(2^{100}) = 2^{\binom{100}{2}} = 2^{4950}$.

• Summing 1 through 100 instead of 1 through 99 (exponent 5050), or reading f(2n) = n f(n) as f(2^n) = 2^n f(2^(n-1)).
• Multiplying the exponents instead of adding them when combining the factors $2^{99} \cdot 2^{98} \cdots$.