Suppose we are given a number "$A$" which is multiple of $\phi(n)$. One can assume factorization to be hard. So you cannot find exact value of $\phi(n)$ from $A$.
Clearly using this we can crack RSA as for any encryption exponent '$e$' we will find its inverse modulo $A$ and that will also we inverse of of '$e$' modulo $\phi(n)$.
But will $A$ factor $n$ ??.
Its looks to me as a natural question on the line of thought of equivalence of RSA and factoring