不妨设n=p1 p2…p36.定义f(n)=([n/5],[2n/5]-[n/5],[3n/5]-[2n/5],[4n/5]-[3n/5],[5n/5]-[4n/5])，若n=0，则f(0)=(0,0,0,0,0)，可忽略.约定∀k∈R (a1+k,a2+k,a3+k,a4+k,a5+k)=(a1,a2,a3,a4,a5 )所以，若a≡b(mod5),则f(a)=f(b)设A=(c1,c2,c3,c4,c5 )=f(n)-[(f(n/p1 )+f(n/p2 )+⋯)]+[(f(n/(p1 p2 ))+⋯)]若n不为5的倍数，设n≡,1/pi ≡ (mod 5)，定义g(x)=f(2x)，则g(x+4)=g(x).则A=g(a0 )-[g(a0+a1 )+g(a0+a2 )+⋯]+[g(a0+a1+a2 )+⋯]取h(x)= (-1)(-1)…(-1)=(x-1)d (x2-1)e (x3-1)f (x4-1)g=-(++⋯)+(+⋯) 于是A中g(t)(t=1,2,3,4)的个数为a0,a0+a1,a0+a2,…中≡(mod 4)的个数，[注：①g(4k+t)视为g(t);② 若a0+ai≡t(mod 4),视作“-1”个”“ ≡t(mod ...