# Find all numbers $n\in \{ 1,...,100 \}$ which have 5 unique divisors.

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.

• @AlexSilva 16 isn't odd though. Nov 21, 2016 at 19:44
• Hint: What kind of numbers have an odd number of divisors? Nov 21, 2016 at 19:46
• I mean just perfect squares. Nov 21, 2016 at 19:48

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.