4. Let $\tau(k)$ denote the number of all positive divisors of a natural number $k$, and suppose the number $n$ is a solution to the equation $\tau(1.6 n)=1.6 \tau(n)$. Determine the value of the ratio $\tau(0.16 n): \tau(n)$.
SOLUTION. If the prime factorization of the number $n$ is $n=\prod_{i=1}^{k} p_{i}^{s_{i}},$ where $p_{1}, \ldots, p_{k}$ are distinct primes and $s_{1}, \ldots, s_{k}$ are non-negative integers, the number of its positive divisors is $\tau(n)=\prod_{i=1}^{k}\left(s_{i}+1\right)$. If $n=2^{\alpha} 5^{\beta} n^{\prime}$ where $\beta \geqq 1$ and $n^{\prime}$ is coprime with $2 \cdot 5$, we can rewrite the given equation as

$$
(\alpha+4) \beta \tau\left(n^{\prime}\right)=\frac{8}{5}(\alpha+1)(\beta+1) \tau\left(n^{\prime}\right),
$$

which, after simplification, gives the equation

$3 \beta(\alpha-4)+8(\alpha+1)=3 \beta(\alpha-4)+8(\alpha-4)+40=(3 \beta+8)(\alpha-4)+40=0$.

From this, it follows that

$$
(3 \beta+8)(4-\alpha)=40=1 \cdot 40 .
$$

However, given that $3 \beta+8 \geqq 11$ and the number $3 \beta+8$ gives a remainder of 2 when divided by 3, the only factorization of the number 40 that satisfies this is

$$
3 \beta+8=20, \quad 4-\alpha=2
$$

thus $\alpha=2, \beta=4, n=2^{2} \cdot 5^{4} n^{\prime}$.

For the ratio $\tau(0.16 n): \tau(n)$, we get

$$
\frac{\tau\left(\frac{4}{25} n\right)}{\tau(n)}=\frac{\tau\left(2^{4} \cdot 5^{2}\right)}{\tau\left(2^{2} \cdot 5^{4}\right)}=\frac{5 \cdot 3}{3 \cdot 5}=1
$$

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