Let $f(m, n)$ denote the minimum for $m, n$. I claim that $f(m, n) = m + n - \gcd(m, n)$ for all $m, n$. Assumptions: Let the cake's mass be $mn$, let $m \le n$, and $n = mk_1 + r_1$. Claim: $f(m, n) = mk_1 + f(m, r_1)$ Proof: Consider the cake split into $m$ slices, each of mass $n$. We want to divide the slices in such a way that they can form $n$ slices, each of mass $m$. It's trivial that each of the final slices must have a mass lesser than or equal to $m$, so the least amount of slices we must cut one slice of mass $n$ into is equal to $k_1 + 1$. To make sure that we have the least amount of slices to form the slices of mass $m$, we can cut each slice of mass $n$ into $k_1$ slices of mass $m$, and $1$ slice of mass $r_1$. Then we will end up with $mk_1$ slices of mass $m$, and $m$ slices of mass $r_1$. We can set the $mk_1$ slices of mass $m$ aside since they are all equal to $m$. We are left with turning $m$ slices of $r_1$ into $r_1$ slices of $m$ using the least amount of slices possible, which will leave us with $f(m, r_1)$ slices. Therefore $f(m, n) = mk_1 + f(m, r_1)$. Claim: $f(m, n) = m + n - \gcd(m, n)$ Proof: If $f(m, n) = m + n - \gcd(m, n)$, then $f(m, r_1) = f(m, n) - mk_1 = m + r - \gcd(m, r_1)$, and if $f(m, r_1) = m + r_1 - \gcd(m, r_1)$ then $f(m, n) = mk_1 + f(m, r_1) = m + n - \gcd(m, n)$ since $\gcd(m, n) = \gcd(m, r_1)$. Therefore $f(m, n) = m + n - \gcd(m, n) \iff f(m, r_1) = m + r_1 - \gcd(m, r_1)$. We can continue similarly to get $f(m, n) = m + n - \gcd(m, n) \iff f(m, r_1) = m + r_1 - \gcd(m, r_1) \iff f(r_2, r_1) = r_2 + r_1 - \gcd(r_2, r_1) \iff ... \iff f(\gcd(m, n), \gcd(m, n)) =\gcd(m, n) + \gcd(m, n) - \gcd(\gcd(m, n), \gcd(m, n)) = \gcd(m, n)$ which is trivial. Therefore our proof is finished.