Is there enough slices 2?

The cake has to be cut to at least 48 pieces if the man wants everyone to have a slice, therefore if there are only 40 guests there will be 8 extra slices. So what will the man do? Simple, he will cut those extra 8 slices into 5 even pieces each so that now he will have 40 small even slices. If 48 guests were to arrive then 40 of them will get a big slice while the other 8 will each get 5 small slices that sum up in size to a big slice, and if 40 guests arrive then each will get one big slice and one small slice. That way all the guests will have a slice in both cases. So to sum it up, the man slices the cake into 40 big slices and 40 small slices so overall there are 80 slices to the cake. In the general case of $m$ or $n$ guests, without loss of generality assume $m<n$ . Then we will slice the cake to $n$ slices. Then, we will take $n-m$ slices and slice each of them to $\frac {LCM(n-m,m)}{n-m}$ slices. This will leave us with $m+(n-m)*\frac{LCM(n-m,m)}{n-m}= \boxed{m+LCM(n-m,m)}$ slices.