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The task is to find all numbers $n\in \{ 1,...,100 \}$ which are divisible by 5 different natural numbers and explain why there is no more numbers than these.
How would I go about that? I did figure out the two numbers: $$divisors(16) = \{ 1,2,4,8,16 \}$$ $$divisors(81) = \{ 1,3,9,27,81 \}$$ The thing I do lack is a good explanation for why it is these two though.
If a number $n$ has a prime factorization $p_1^{a_1}p_2^{a_2} \dots p_n^{a_n}$, then the number of positive divisors of $n$ is given by $(a_1+1)(a_2+1)\dots(a_n+1)$.
Since $5$ is a prime number, we must have $a_1=4$, and $a_2,a_3, \dots ,a_n =0$.
i.e. $n= (p_1)^4$ for some prime number $p_1$.
For $p_1=2$, $n=16$ and for $p_1=3$, $n=81$. For bigger $p_1$, $n>100$.
Note that both numbers you have found are powers of primes.
This is important, since every number is divisible by 1 and itself.
That leaves place for 3 divisors. Only one can be a prime, however. Because if there were two different primes $x$ and $y$ ($x \ne y$) then
$xy$ is also a divisor, but not your number yet, since you need one more divisor.
Because your number has only two prime divisors
$xxy$ is a divisor
$xyy$ is also a divisor. $xxy \ne xyy$ because $x\ne y$, therefore neither of those can be your number.
That is at least 5+2=7 divisors.
Therefore, since $x \ne y$ and your number is bigger than $x$ or $y$, only one of those can be prime. This implies that only numbers that are in form $p^4$, where $p$ is a prime, have 5 divisors.
$p^4 < 100$ only for primes $p = 2,3$ as you have already found out.
