Modulo problem

OK! Here is my best understanding of this:

A=1 + 3 + 5 + 7 + 9 + 11 + 13 + 15=64 sum of all  invertible integers mod 2^4.
B=2 + 4 + 6 + 8 + 10 + 12 + 14      =56 sum of all non-invertible integers mod 2^4.
A - B =64 - 56 = 8

Note: Somebody should check this. heureka maybe?

For each positive integer n,
the set of integers \(\{0,\ 1,\ 2,\ \ldots \,\ n-1 \}\) is known as the residue system modulo n.
Within the residue system modulo \(2^4\),
let A be the sum of all invertible integers modulo \(2^4\) and
let B be the sum all of non-invertible integers modulo \(2^4\).
What is A-B?

The integer s of the set is invertible, if \(gcd(2^4,s)=1\)

\(\begin{array}{|c|c|c|l|} \hline \text{set of integers} & gcd(2^4,s) & \text{not invertible} & \text{modulo inverse} \\ \hline 0 & 16 & \checkmark \\ \hline 1 & 1 && 1^{-1} \pmod{16} = 1 \quad |\quad 1\cdot 1 \equiv 1 \pmod{16} \\ \hline 2 & 2 & \checkmark \\ \hline 3 & 1 && 3^{-1} \pmod{16} = 11 \quad |\quad 3\cdot 11 \equiv 1 \pmod{16} \\ \hline 4 & 4 & \checkmark \\ \hline 5 & 1 && 5^{-1} \pmod{16} = 13 \quad |\quad 5\cdot 13 \equiv 1 \pmod{16} \\ \hline 6 & 2 & \checkmark \\ \hline 7 & 1 && 7^{-1} \pmod{16} = 7 \quad |\quad 7\cdot 7 \equiv 1 \pmod{16} \\ \hline 8 & 8 & \checkmark \\ \hline 9 & 1 && 9^{-1} \pmod{16} = 9 \quad |\quad 9\cdot 9 \equiv 1 \pmod{16} \\ \hline 10 & 2 & \checkmark \\ \hline 11 & 1 && 11^{-1} \pmod{16} = 3 \quad |\quad 11\cdot 3 \equiv 1 \pmod{16} \\ \hline 12 & 4 & \checkmark \\ \hline 13 & 1 && 13^{-1} \pmod{16} = 5 \quad |\quad 13\cdot 5 \equiv 1 \pmod{16} \\ \hline 14 & 2 & \checkmark \\ \hline 15 & 1 && 15^{-1} \pmod{16} = 15 \quad |\quad 15\cdot 15 \equiv 1 \pmod{16} \\ \hline \hline \end{array}\)

\( A = 1+3+5+7+9+11+13+15 = \mathbf{64} \\ B = 0+2+4+6+8+10+12+14 = \mathbf{56}\)