為甚麼7的倍數判斷是3個一組，然後奇數節減偶數節再除以7啊?求解拜託?

Sol
100000=14285*7+5
10000=1428*7+4
1000=142*7+6
100=14*7+2
10=7+3
ABCDEF=A*10^5+B*10^4+C*10^3+D*10^2+E*10+F
=(14285*A+1428*B+142*C+14*D+E)*7+5*A+4*B+6*C+2*D+3*E+F
=M*7+5*A+4*B+6*C+2*D+3*E+F
ABC=100*A+10*B+C
=(14*A+B)*7+2*A+3*B+C
=N*7+2*A+3*B+C
DEF=100*D+10*E+F
=P*7+2*D+3*E+F
So
DEF－ABC
=(P*7+2*D+3*E+F)－(N*7+2*A+3*B+C)
=P*7+2*D+3*E+F－N*7－2*A－3*B－C
=P*7+2*D+3*E+F－N*7－7*A－7*B－7*C+5*A+4*B+6*C
=(P－N－A－B－C)*7+2*D+3*E+F+5*A+4*B+6*C
So
ABCDEFD=W*7+(DEF－ABC)
ABCDEF－DEF+ABC=W*7