Yes we can factor an RSA modulus $n$ given $n$ and $k \cdot \phi(n)$, including where $k$ is a (reasonably) large prime. We just use $f\gets k \cdot \phi(n)+1$ instead of $f\gets e\,d-1$ in the algorithm of this answer.

The algorithm is heuristic, and I do not claim a rigorous proof of the distribution of the runtime. Also, a larger $k$ tends to make it more costly. But in practice it works fine for $k$ of size comparable to $n$.

Yes we can factor an RSA modulus $n$ given $n$ and $k \cdot \phi(n)$, including where $k$ is a (reasonably) large prime. We just use $f\gets k \cdot \phi(n)$ instead of $f\gets e\,d-1$ in the algorithm of this answer.

The algorithm is heuristic, and I do not claim a rigorous proof of the distribution of the runtime. Also, a larger $k$ tends to make it more costly. But in practice it works fine for $k$ of size comparable to $n$.

Yes one can factor an RSA modulus $n$ if you know $k \cdot \phi(n)$, including where $k$ is a (reasonably large) large prime. One just uses $k \cdot \phi(n)$ where there is $e\,d$ in the algorithm of this answer.

The algorithm is heuristic, and I do not claim a rigorous proof of the distribution of the runtime. Also, a larger $k$ tends to make it more costly. But in practice it works fine for $k$ of size comparable to $n$.

Yes we can factor an RSA modulus $n$ given $n$ and $k \cdot \phi(n)$, including where $k$ is a (reasonably) large prime. We just use $f\gets k \cdot \phi(n)+1$ instead of $f\gets e\,d-1$ in the algorithm of this answer.

The algorithm is heuristic, and I do not claim a rigorous proof of the distribution of the runtime. Also, a larger $k$ tends to make it more costly. But in practice it works fine for $k$ of size comparable to $n$.