Statement Every even perfect number N=2p−1(2p−1)N=2^{p-1}(2^p-1)N=2p−1(2p−1) with ppp an odd prime satisfies N≡4(mod12)N\equiv 4\pmod{12}N≡4(mod12). Verify this for N=8128N=8128N=8128. Solution 8128=12⋅677+4,8128=12\cdot 677+4,8128=12⋅677+4, hence 8128≡4(mod12).8128\equiv \boxed{4}\pmod{12}.8128≡4(mod12).