In base $b$, let $G_n$ be the set of $n$-digit integers, be the integers mod $b^n$. Then we have an exact sequence $$0\rightarrow G_1\rightarrow G_n\rightarrow G_{n-1}\rightarrow 0$$

For $n-1$ digit numbers, the leftmost carry digit is the two-cocycle associated to this group extension and therefore satisfies the cocycle condition $$c(x,y)+c(x+y,z)=c(y,z)+c(x,y+z)$$ In other words, the sum of the leftmost carry digits does not depend on the order of the summands.

After reducing mod $b^k$, the same argument holds for any other carry digit.

In base $b$, let $G_n$ be the set of $n$-digit integers, thought of as the integers mod $b^n$. Then we have an exact sequence $$0\rightarrow G_1\rightarrow G_n\rightarrow G_{n-1}\rightarrow 0$$

For $n-1$ digit numbers, the leftmost carry digit is the two-cocycle associated to this group extension and therefore satisfies the cocycle condition $$c(x,y)+c(x+y,z)=c(y,z)+c(x,y+z)$$ In other words, the sum of the leftmost carry digits does not depend on the order of the summands.

After reducing mod $b^k$, the same argument holds for any other carry digit.

In base $b$, let $G_n$ be the set of $n$-digit integers, be the integers mod $b^n$. Then we have an exact sequence $$0\rightarrow G_1\rightarrow G_n\rightarrow G_{n-1}\rightarrow 0$$

For $n-1$ digit numbers, the leftmost carry digit is the two-cocycle associated to this group extension and therefore satisfies the cocycle condition $$c(x,y)+c(x+y,z)=c(y,z)+c(x,y+z)$$ In other words, the leftmost carry digit does not depend on the order of the summands.

After reducing mod $b^k$, the same argument holds for any other carry digit.

In base $b$, let $G_n$ be the set of $n$-digit integers, be the integers mod $b^n$. Then we have an exact sequence $$0\rightarrow G_1\rightarrow G_n\rightarrow G_{n-1}\rightarrow 0$$

For $n-1$ digit numbers, the leftmost carry digit is the two-cocycle associated to this group extension and therefore satisfies the cocycle condition $$c(x,y)+c(x+y,z)=c(y,z)+c(x,y+z)$$ In other words, the sum of the leftmost carry digits does not depend on the order of the summands.

After reducing mod $b^k$, the same argument holds for any other carry digit.