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How many natural numbers $n$ have such a property that out of all the positive divisors of number $n$, which are different from both $1$ and $n$, the greatest one is $15$ times greater than the smallest one?

After trying out different combinations, one will see, that the only ones that seem to work, are for $n=60$ and $n=135$.

Question: How to prove that there can be no others?

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Hint: If $a$ and $b$ are, respectively, the smallest and the greatest divisors of $n$ (without $1$ and $n$ itself), then

$$a·b=n\iff a·15a=15a^2=n\implies 3\mid n$$

Can you end it now?

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Hint

Note that $n$ must be a multiple of $15.$ Thus, in particular, it is a multiple of $3.$ So, the smallest divisor (different from $1$) of $n$ is $2$ or $3.$ Thus, the biggest divisor (different from $n$) of $n$ is $30$ or $45.$

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Let's write down a factorization of $n$ $$n = p_1^{k_1}p_2^{k_2}...p_m^{k_m}$$ where $p_1...p_m$ are in increasing order.

Clearly the the smallest divisor of $n$ is $p_1$, and the greatest is $p_1^{k_1 - 1}p_2^{k_2}...p_m^{k_m} = D$.

So we have $$15p_1 = p_1^{k_1 - 1}p_2^{k_2}...p_m^{k_m}$$ $$3*5*p_1 = p_1^{k_1 - 1}p_2^{k_2}...p_m^{k_m}$$ Both sides are factorizations of $D$, therefore (by fundamental theorem of arithmetic) they are the same.

Clearly $p_1$ can not be greater than $3$ (otherwise it is not the smallest prime in factorzation of $D$), thus $p \in \{2, 3\}$

If $p = 2$ we have $D = 2^{1} * 3^{1}* 5^{1}$ and $n = 2^{2} * 3^{1}* 5^{1} = 60$

If $p = 3$ we have $D = 3^{2}* 5^{1}$ and $n = 3^{3}* 5^{1} = 135$

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