Question

二.三个2002位数的运算99...9*88...8/66

二.三个2002位数的运算99...9*88...8/66...6的结果中为何 有相邻的2001个3 详解此题

Answer 1

99...9
*
88...8
/
66...6
（分子分母约掉个2）
=
99...9
*
44...4
/
33...3
=
3
*
44...4
三四一十二，进位，所以
=
1
33...3
2
（共2003位，故中间有2001个3）
或：
99...9
=
100...0
-
1
=
10^2002
-
1
88...8
=
(8/9)
*
99...9
=
(8/9)
*
(10^2002
-
1)
66...6
=
(6/9)
*
99...9
=
(2/3)
*
(10^2002
-
1)
　99...9
*
88...8
/
66...6
=
(10^2002
-
1)
*
(8/9)
*
(10^2002
-
1)
/
[(2/3)
*
(10^2002
-
1)]
=
(4/3)
*
(10^2002
-
1)
=
(4/3)
*
99...9
=
4
*
33..3
三四一十二，进位，所以为：
133...32