China Mathematical Olympiad

The answer is YES.
From the definition of $\omega$ and $\Omega$, we have
$$\omega(ab) \le \omega(a) + \omega(b), \qquad \textcircled{1}$$
$$\Omega(ab) = \Omega(a) + \Omega(b), \qquad \textcircled{2}$$
for any positive integers $a, b$. Given a fixed positive integer $k$ and positive real numbers $\alpha, \beta$, we take a positive integer $m > (\omega(k) + 1)\alpha$. As there are infinitely many prime numbers, we can take a sufficiently large prime $p$ such that $\frac{\Omega(k)+1}{p^m} + \log_p 2 < \beta$, and take $m$ pairwise distinct prime numbers $q_1, q_2, \dots, q_m$ that are all greater than $p$. We will show that $n = 2^{q_1 q_2 \cdots q_m} k$ has the desired property.

First, we prove $\frac{\omega(n+k)}{\omega(n)} > \alpha$. Let $n_1 = \frac{n+k}{k} = 2^{q_1 q_2 \cdots q_m} + 1$. As $q_1, q_2, \dots, q_m$ are all odd prime numbers, $2^{q_i} + 1 \mid n_1$ when $1 \le i \le m$, then $d_i = \frac{2^{q_i} + 1}{3}$ is an integer greater than 1.

Note that
$(2^r - 1, 2^s - 1) = 2^{(r,s)} - 1$ for all positive integers $r, s$, ③
and $(q_i, q_j) = 1\ (i \neq j)$, we have
$$\begin{aligned} (d_i, d_j) &= \frac{1}{3}(2^{q_i} + 1, 2^{q_j} + 1) \le \frac{1}{3}(2^{2q_i} - 1, 2^{2q_j} - 1) \\ &= \frac{2^{(2q_i, 2q_j)} - 1}{3} = \frac{2^2 - 1}{3} = 1. \end{aligned}$$
$d_1, d_2, \dots, d_m$ are the pairwise coprime factors of $n_1$, and each of them is greater than 1. Hence, $\omega(n_1) \ge m$. From ① and the choice of $m$, we have
$$\frac{\omega(n+k)}{\omega(n)} \ge \frac{\omega(n_1)}{\omega(n)} \ge \frac{\omega(n_1)}{\omega(k)+1} \ge \frac{m}{\omega(k)+1} > \alpha.$$

Next, we prove $\frac{\Omega(n+k)}{\Omega(n)} < \beta$. As $q_1 q_2 \cdots q_m$ is an odd number and cannot be divided by 3, we have $n_1 = 2^{q_1 q_2 \cdots q_m} + 1 \equiv \pm 3 \pmod{9}$, that is, $3 \nmid n_1$. Suppose $q$ is a prime factor of $\frac{n_1}{3}$ and $q \le p$, then
$$2^{2q_1 q_2 \cdots q_m} - 1 = (2^{q_1 q_2 \cdots q_m} - 1) \cdot n_1 \equiv 0 \pmod{q}.$$
From the Fermat's Little Theorem, $2^{q-1} \equiv 1 \pmod{q}$. From ③, $q \mid 2^{(2q_1 q_2 \cdots q_m, q-1)} - 1$. From $(q-1, 2q_1 q_2 \cdots q_m) = (q-1, 2) \le 2, q-1 < p < q_i\ (i = 1, 2, \dots, m)$, hence $q \mid 2^2 - 1, q = 3$. This contradicts that $\frac{n_1}{3}$ is not a multiple of 3. Therefore, each prime factor of $\frac{n_1}{3}$ is larger than $p$. So $\frac{n_1}{3} > p^{\Omega(n_1/3)}$. From ② and the choice of primes $p$ and $q_1, q_2, \dots, q_m$, we have
$$\begin{align*} \Omega(n+k) &= \Omega(k) + \Omega(3) + \Omega\left(\frac{n_1}{3}\right) < \Omega(k) + 1 + \log_p\left(\frac{n_1}{3}\right) \\ &< \Omega(k) + 1 + \log_p(n_1 - 1) \\ &= \Omega(k) + 1 + q_1 q_2 \cdots q_m \log_p 2, \end{align*}$$
$$$$
$\text{(n+k) (n) < (k)}$ + 1 + q_1 q_2 $\cdots$ q_m $\log_p$ 2}{q_1 q_2 $\cdots$ q_m} < $\text{(k)}$ + 1}{p^m} + $\log_p$ 2 < $\beta. \quad \square$