[ $\quad$ Number of divisors and their sum of a number [Fundamental Theorem of Arithmetic. Factorization into prime factors ]

a) Find the number $k$, which is divisible by 2 and 9 and has exactly 14 divisors (including 1 and $k$ ).

b) Prove that if 14 is replaced by 15, the problem will have multiple solutions, and if 14 is replaced by 17, there will be no solutions at all.
Let $M=p_{1}^{\alpha_{1}} \ldots p_{r}^{\alpha_{r}}$ be the prime factorization of the number $M$; for definiteness, we will assume that $\alpha_{1} \geq \alpha_{2} \geq \ldots \geq \alpha_{r} \geq 1$. According to the formula from problem $\underline{60537}$ a) the total number of divisors of $M$ is $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots \left(\alpha_{r}+1\right)$.

a) $14=7 \cdot 2$, so $r=2, \alpha_{1}=6, \alpha_{2}=1$. We know that 3 is included in the factorization of $M$ with an exponent greater than 1, and 2 with an exponent greater than 0. Therefore, $M=3^{6} \cdot 2=1458$.

b) For 15 divisors, possible answers are $M=3^{2} \cdot 2^{4}$ and $M=3^{4} \cdot 2^{2}$, and 17 divisors can only be had by numbers of the form $M=p^{16}$, where $p$ is a prime.

## Answer

a) 1458.