Question

试确定(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)的末位数字.

Answer 1

(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)

=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)/(2-1)

=(2^2-1)(2^2+1).(2^64+1)

=...

=(2^64-1)(2^64+1)

=(2^128-1)

2^1=2

2^2=4

2^3=8

2^4个位数是6

2^5个位数是2

所以个位数是4个一循环

所以2^128的个位数=2^4的个位数=6

即：2^128-1的个位上是：6-1=5