- Compute all the prime factors of $k$. It would take $O(\sqrt{k})$ time $\approx$ $10^5$ time. Note that, there could be at most $10$ different prime factors of $k$ since considering the $10$ smallest prime numbers gives the product: $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29$ which exceeds $10^9$. Let us call the prime factors of $k$ as: $f_{1}, \dotsc,f_{10}$ with multiplicities $m_{1}, \dotsc,m_{10}$, respectively. The sum of multiplicities (or total number of prime factors) $\sum_{i = 1}^{10}m_{i}$ could be at most $30$ since $2^{30}$ exceeds $10^{9}$.
- Now, you also need to factorize the numbers given in array $a$. However, you do not need to obtain all its factors. You just need to obtain the factors which are among $f_{1},\dotsc,f_{10}$ since other prime factors are irrelevant for $\prod\limits_{x\in m}x \equiv 0\mod k$. Therefore, just check that if an array entry $a[j]$ is divisible by some factors $f_{i}$ (obtained in step 2) and obtain its multiplicity in $a[j]$. Let $m_{i}^{j}$ denote the multiplicity of factor $f_{i}$ for an array entry $a[j]$. Computing all $m_{i}^{j}$ would take at most $\sum_{i = 1}^{10}m_{i} \cdot n = 30 \cdot n$ time $\approx 10^{6}$ time.
- Now, you just need a Dynamic Programming approch to obtain the minimum value on size of $m$. Let $A$ denote a matrix of dimension $(m_{1}+1)$x $(m_{2}+1)$ x $\dotsc$x $(m_{10}+1)$ x $n$ such that an entry $A[j_{1}][j_{2}] \dotsc[j_{10}][t]$ denote the minimum size value of $m$ such that $\prod\limits_{x\in m}x \equiv 0\mod k'$ for $m \subseteq \{a_{1},\dotsc,a_{t}\}$ and $k' = \prod_{i = 1}^{10} j_{i} \cdot f_{i}$. In otherwords, $k'$ is composed of all possible combinations of the factors $f_{1}, \dotsc,f_{10}$. Therefore $k'$ can take at most $(m_{1}+1) \cdot (m_{2}+1) \dotsc (m_{10}+1) \approx 2^{10} \approx 10^{3}$ different values. This bound is based on the assumption that all $10$ factors are distinct. If all the factors are the same the we have $k' \leq 30$ since $2^{30}$ exceeds $10^9$. So I believe $k'$ would take at most $\approx 2 \cdot 2^{10} \approx 2 \cdot 10^3$ different values.
Now, assuming 1 based indexing for the matrix $A$, we define the DP steps as follows.
The Main DP step:
$$
A[j_{1}][j_{2}]\dotsc[j_{10}][t+1] = \min \left\{\begin{array}{lr}
A[j_{1}][j_{2}]\dotsc[j_{10}][t], & \textrm{do not pick $a[t+1]$}\\
1 + A[j_{1}-m_{1}^{t}][j_{2}-m_{2}^{t}]\dotsc[j_{10}-m_{10}^{t}][t], & \textrm{pick $a[t+1]$}\\
\end{array}\right.
$$
The Base Step:
$A[j_{1}][j_{2}]\dotsc[j_{10}][1] = 1$ if $j_{i} \leq m_{i}^{1}$ for all $1 \leq i \leq 10$. And $A[j_{1}][j_{2}]\dotsc[j_{10}][1] = \infty$ otherwise. Here $\infty$ means that it is not possible to satisfy the constraint $\prod\limits_{x\in m}x \equiv 0\mod k'$ given $m \subseteq \{a_{1} \}$ and $k' = \prod_{i = 1}^{10} j_{i} \cdot f_{i}$.
The Required Ouput:
The output of the DP procedure would be the solution: $A[m_{1}][m_{2}]\dotsc[m_{10}][n]$ which naturally denote the minimum size value of $m$ such that $\prod\limits_{x\in m}x \equiv 0\mod k'$ for $m \subseteq \{a_{1},\dotsc,a_{n}\}$ and $k' = k = \prod_{i = 1}^{10} m_{i} \cdot f_{i}$.
Running Time Analysis of DP:
The size of the matrix $A$ is $(\textrm{number of possible value of $k'$}) \cdot n \approx 2 \cdot 10^3 \cdot 10^4 \approx 2 \cdot 10^{7}$. Computing each entry takes 2 or 3 operations. Therefore total running time would be $\approx 6 \cdot 10^7$.
Thus the overall running time is $\approx 10^5 + 10^6 + 6 \cdot 10^ 7 \approx 10^8$. I have done the loose analysis, so it should be doable in 3 sec.
Total Space Complexity $\approx $ size of the matrix $A$ $\approx 2 \cdot 10^7$
