2004 AMC 10A Problem 24

Start with small powers of $2$ to see what the rule does.

1. Rule (ii) with $n = 1$: $f(2) = 1 \cdot f(1) = 1$.
2. With $n = 2$: $f(4) = 2 \cdot f(2) = 2$.
3. With $n = 4$: $f(8) = 4 \cdot f(4) = 4 \cdot 2 = 8$.
4. With $n = 8$: $f(16) = 8 \cdot f(8) = 8 \cdot 8 = 64$.

Write these as powers of $2$:

5. Why the exponents $0, 1, 3, 6$? Each step uses $n = 2^{k-1}$, so it multiplies by $2^{k-1}$:
$$f(2^k) = 2^{k-1} \cdot f(2^{k-1}).$$
Multiplying powers of $2$ adds their exponents. So the exponent of $f(2^k)$ is $0 + 1 + 2 + \cdots + (k-1)$. For $k = 4$ that is $0 + 1 + 2 + 3 = 6$, which matches.
6. For $k = 100$, the exponent is
$$0 + 1 + 2 + \cdots + 99 = \frac{99 \times 100}{2} = 4950.$$

So $f(2^{100}) = 2^{4950}$.

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

A rule that links $f(2n)$ to $f(n)$ lets you climb the powers of $2$ one step at a time, multiplying by one new factor at each step. Working out the first few values shows the pattern, and a table of exponents makes it plain that you are adding $0 + 1 + 2 + \cdots$. Be careful where the sum ends: to reach $2^{100}$ the last factor is $2^{99}$, not $2^{100}$.

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