IS Array of Positive Integers Always 'Equalizable' With This Strategy? A little bit of history first. I have asked this question here some days ago. Quoting from there:


  An array $A$ is given that has only positive integers in it. The objective is to equalize the array in minimum possible steps. In every step all but one element are incremented by a fixed step size. The possible step sizes are given in another array $D$.For example, if $A=\{2,2,3,7\}$ and $D=\{1,2,5\}$ we can equalize $A$ in two steps, first adding $A=\{5,5,5,0\}$ to get $A=\{7,7,8,7\}$ and then adding $\{1,1,0,1\}$ to get $A=\{8,8,8,8\}$.


If $A$ has just two elements, this reduces to something like a change-making problem. If $\Delta=min(A[0],A[1])$, if I can make change for $\Delta$ using only the given steps from $D=\{1,2,5\}$, I can equalize the array by holding the larger element constant while incrementing the smaller element by the same values that I used to make change for $\Delta$. However,if it's not possible to make change for delta, it's not clear if the two elements can converge two some larger value following this strategy. 

I would like to know if this strategy of withholding one element and incrementing the rest by a chosen 'delta' will always make all elements equal in some finite number of steps.
If the GCD of the array D is larger than 1, then this cannot always be done. The simplest example is $A = \{1,2\},D=\{2\}$. In contrast, if the GCD of the array D is 1, then it is known that large enough integers can be represented as positive linear combinations of elements of D. In particular, there is an integer $N$ such that $N-a$ can be represented for all $a \in A$. Summing all these representations will equalize the array.

For example, when $A = \{2,3,7\}$ and $D = \{1,2,5\}$, we can add $\{1,0,0\}$ once and $\{1,1,0\}$ four times, reflecting the representations $7-2=5 \times 1$, $7-3=4 \times 1$, $7-7 = 0 \times 1$. Even when $D = \{2,5\}$ this has a solution: using $11 = 3 \times 2 + 1 \times 5$, $10 = 2 \times 5$, $6 = 3 \times 2$, we can add $\{2,0,2\}$ three times, $\{5,5,0\}$ once, and $\{0,5,0\}$ once to get $\{13,13,13\}$.