C/1 Find all natural numbers $n$ such that when we multiply all divisors of $n$, we will obtain $10^{9}$. Prove that your number(s) $n$ works and that there are no other such numbers.
(Note: A natural number $n$ is a positive integer; i.e., $n$ is among the counting numbers $1,2,3$, $\ldots$ A divisor of $n$ is a natural number that divides $n$ without any remainder. For example, 5 is a divisor of 30 because $30 \div 5=6$; but 5 is not a divisor of 47 because $47 \div 5=9$ with remainder 2. In this problem we consider only positive integer numbers $n$ and positive integer divisors of $n$. Thus, for example, if we multiply all divisors of 6 we will obtain 36.)
C/1 Solution 1: Since the prime factorizaton of $10^{9}=2^{9} \cdot 5^{9}$, the prime divisors of our natural number $n$ are exactly 2 and 5 ; i.e., $n=2^{a} \cdot 5^{b}$ for some integer exponents $a \geq 1$ and $b \geq 1$.
When $a=b=2$, we get the number $n=2^{2} \cdot 5^{2}=100$, which actually works: the product of its divisors is:
$$
\begin{aligned}
1 \cdot 2 \cdot 4 \cdot 5 \cdot 10 \cdot 20 \cdot 25 \cdot 50 \cdot 100 & =(1 \cdot 100) \cdot(2 \cdot 50) \cdot(4 \cdot 25) \cdot 10 \\
& =100^{4} \cdot 10=10^{2} \cdot 10^{2} \cdot 10^{2} \cdot 10^{2} \cdot 10=10^{9} .
\end{aligned}
$$

In fact, $n=100$ is the only number that works. Indeed, if in $n=2^{a} \cdot 5^{b}$ we have, say, $b \geq 3$, we will end up with too many $5 \mathrm{~s}$ in the product of all divisors! To see this, note that we will have at least the divisors $5,25,125,2 \cdot 5,2 \cdot 25$, and $2 \cdot 125$. This yields at least $1+2+3+1+2+3=12$ fives in $10^{9}$, but we have only 9 such $5 \mathrm{~s}$, a contradiction. Similarly, in $a \geq 3$, we will end up with at least 12 twos in $10^{9}$, which is also a contraduction. We conclude that $a \leq 2$ and $b \leq 2$.
On the other hand, if, say $b=1$, we will get too few fives in the product! Indeed, $n=2^{a} \cdot 5$ with $a \leq 2$, so that the $5 \mathrm{~s}$ in the product will participate in at most the following divisors: $5,2 \cdot 5,2^{2} \cdot 5$; i.e., only at most 3 fives, but we need 9 fives in $10^{9}$, a contradiction! Similarly, we will get too few 2 s in the product if $a=1$. We conclude that $a=b=2$ and the only number that works is $n=2^{2} \cdot 5^{2}=100$.

Solution 2: If we arrange all divisors of $n$ in increasing order, $d_{1}<d_{2}<d_{3}<\ldots<d_{k}$, and multiply the first and the last divisor, the second and the last but one divisor, etc., we will always obtain $n$ (why?). This means that if we multiply the product $\Pi=d_{1} \cdot d_{2} \cdot d_{3} \cdots d_{k}$ by itself, we will obtain:
$$
\Pi^{2}=\left(d_{1} d_{k}\right)\left(d_{2} d_{k-1}\right) \cdots\left(d_{k} d_{1}\right)=\underbrace{n \cdot n \cdots n}_{k}=n^{k},
$$
where $k$ is the number of divisors of $n$.
We are given that the product $\Pi=10^{9}$, so we conclude that $\left(10^{9}\right)^{2}=n^{k}$, i.e., $n^{k}=10^{18}$. From here it is easy to see that $n$ must be a power of 10 ; i.e., $n=10^{c}$ for some natural $c$ such that $\left(10^{c}\right)^{k}=10^{18}$. Thus, $10^{c k}=10^{18}, c k=18$, and $c$ is a divisor of 18 . One may think that we need now check all cases for $c$, but there are faster ways to proceed. You may want to check that $c=2$ works ( $n=100$ works as in Solution 1), argue that $10^{3}$ and higher powers of 10 will yield larger than $10^{9}$ products $\Pi$ and that lower powers of 10 (i.e., $10^{1}$ ) will yield smaller than $10^{9}$ products $\Pi$.
Alternatively, note that $n=10^{c}=2^{c} \cdot 5^{c}$ means that $n$ will have $k=(c+1) \cdot(c+1)$ divisors of $n$ (why?), and hence $c k=18$ becomes $c(c+1)(c+1)=18$, which has only one natural solution of $c=2$ (why?).
Summarizing, $n=10^{c}=10^{2}=100$ is the only possible solution.