解(1)τ(5246731)=0+1+1+0+0+4+6=12;
(2) τ (314265)=0+1+0+2+0+1=4;
(3) τ (654321)=0+1+2+3+4+5=15;
(4) τ (1(2k)2(2k-1) ···(k-1)(k+2)k(k+1))
+(k-1)=k(k-1);
(5) τ (369···(3k)258···(3k-1)147···(3k-2))
=0+0+···+0+k+(k-1)+ ···+1+2k+2(k-1)
+···+2
=3×(k+(k-1)+ ···+1)
=
k(k+1)