Let $a_1, a_2, \ldots$ be a sequence with the following properties.

(i) $a_1 = 1$, and

(ii) $a_{2n}=n\cdot a_n$ for any positive integer $n$.

What is the value of $a_{2^{100}}$?
Note that

\begin{align*}
a_{2^1} &= a_2 = a_{2\cdot1} = 1\cdot a_1 = 2^0\cdot 2^0 = 2^0,\\
a_{2^2} &= a_4 = a_{2\cdot2} = 2\cdot a_2 = 2^1\cdot 2^0 = 2^1,\\
a_{2^3} &= a_8 = a_{2\cdot4} = 4\cdot a_4 = 2^2 \cdot 2^1 = 2^{1+2},\\
a_{2^4} &= a_{16} = a_{2\cdot8} = 8\cdot a_8 = 2^3\cdot 2^{1+2} = 2^{1+2+3},
\end{align*}and, in general, $a_{2^n} = 2^{1+2+\cdots+(n-1)}$.  Because $$1+2+3+\cdots+(n-1) =
\frac{1}{2}n(n-1),$$we have $a_{2^{100}} = 2^{(100)(99)/2} = \boxed{2^{4950}}$.
2^{4950}
Level 3
Intermediate Algebra
Problem: Let $a_1, a_2, \ldots$ be a sequence with the following properties.

(i) $a_1 = 1$, and

(ii) $a_{2n}=n\cdot a_n$ for any positive integer $n$.

What is the value of $a_{2^{100}}$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of $a_{2^{100}}$, we start by applying the recursive definition given in property (ii). 

First, we need to express $2^{100}$ in terms of the sequence's properties. We can write $2^{100}$ as $2 \cdot 2^{99}$. By applying the recursive formula, we have:

1. For $n = 2^{99}$, we can express $a_{2^{100}}$ as:
   $$ a_{2^{100}} = a_{2 \cdot 2^{99}} = 2^{99} \cdot a_{2^{99}}. $$

Next, we need to find $a_{2^{99}}$. We can apply the same reasoning:

2. For $n = 2^{98}$, we have:
   $$ a_{2^{99}} = a_{2 \cdot 2^{98}} = 2^{98} \cdot a_{2^{98}}. $$

Continuing this process, we can see a pattern emerging. We can express $a_{2^{98}}$ in a similar manner:

3. For $n = 2^{98}$:
   $$ a_{2^{98}} = a_{2 \cdot 2^{97}} = 2^{97} \cdot a_{2^{97}}. $$

This pattern continues down to $a_1$:

4. Eventually, we reach:
   $$ a_{2^1} = a_{2 \cdot 1} = 1 \cdot a_1 = 1. $$

Now, we can substitute back up through the chain of equations:

- From $a_{2^{99}} = 2^{98} \cdot a_{2^{98}}$, we substitute $a_{2^{98}}$.
- From $a_{2^{98}} = 2^{97} \cdot a_{2^{97}}$, we substitute $a_{2^{97}}$.
- This continues until we reach $a_1 = 1$.

The general form we derive is:

$$ a_{2^k} = 2^{k-1} \cdot a_{2^{k-1}}. $$

By continuing this substitution, we find:

$$ a_{2^{100}} = 2^{99} \cdot a_{2^{99}} = 2^{99} \cdot (2^{98} \cdot a_{2^{98}}) = 2^{99} \cdot 2^{98} \cdot (2^{97} \cdot a_{2^{97}}) = \ldots = 2^{99 + 98 + 97 + \ldots + 1} \cdot a_1. $$

The sum $99 + 98 + 97 + \ldots + 1$ can be calculated using the formula for the sum of the first $n$ integers:

$$ S = \frac{n(n + 1)}{2}, $$

where $n = 99$. Thus,

$$ S = \frac{99 \cdot 100}{2} = 4950. $$

Since $a_1 = 1$, we conclude:

$$ a_{2^{100}} = 2^{4950}. $$

Therefore, the final answer is $2^{4950}$.<|END_OF_LATENT|> The final answer is $2^{4950}$.