Hi, I'd like to share my math understanding of this problem.

This problem is actually a Prime factorization problem.

let's say the target number is 12.

you have two ways to get 12 with minimum steps.

12 = 6+6; 6=3+3; 3= 1+1+1. total steps is 7. OR

12= 4+4+4; 4=2+2; 2=1+1. total steps is also 7.

which means, whatever you decide to paste, as long as you paste a prime number times, the result will be optimum and this prime number is in its prime factorization.

Here is the prove:

A = a_1*a_2*...a_n. where a1 to an are all prime number(may have duplicates).

let A_m = a_1*...*a_m.
B_m is the total steps to get A_m.
B_(m+1)=B_m+a_(m+1). copy all than paste (a_(m+1)-1) times, total a_(m+1) times.
so, as you can see, from B_m to B_m+1 is just add some constant number, there is NO addition between different .
B = a_1+a_2...+a_n.
the result is a constant number correspond to A itself. So, the sequence of how you use this prime number will not affect the result.
still use 12 as example: 12 = 2*2*3. 7=2+2+3.

As for the part why this is optimum

if you just do a factorization:

A=a_1*a_2... c_1...a_n, where c_1=a_ka_(k+1)*...*a_(k+l),

c1>=a_k+a_(k+1)+...+a_(k+l)