If $n$ is even, then we write it as: $n = (2n)-n$ as both $(2n)$ as $n$ have no of same prime factors as $n$. If $n$ is odd, let $d$ be the smallest odd prime that does not divide $n$. Then: $n = dn - (d-1)n$. Note that: $dn$ has exactly one more prime factor than $n$. Since $n, d$ is odd, $(d-1)$ is divisible by $2$ and rest of the prime factors of $d-1$ also also prime factors of $n$ as $d$ is the smallest odd prime that doesnt divide $n$. Hence both $nd$ and $(d-1)n$ have same number of prime factors.