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.

(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

(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)=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=(2^8-1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=(2^16-1)(2^16+1)(2^32+1)(2^64+1)=(2^32-1)(2^32+1)(2^64+1)=(2^64-1)(2^64+1)=2^128-1

2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
所以每4次，末位分别是2、4、8、6的循环，因为128=4*32，所以2^128的末位为6，所以原式的末位为6-1=5

3×5×9×17×33×65×129

末位为5