Question 72
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, .... 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 .)
(Note: A natural number $n$ is a positive integer; i.e., $n$ is among the counting numbers 1, 2, 3, .... 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 .)