Question

(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1的末位数字. 怎么做?(详细过程)

Answer 1

(2^2+1) 等于5，其他各项均为奇数。
故(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1) 个位数等于 5.
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1 个位数等于 6.

Answer 2

(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
先乘以(2^2-1) 再除以(2^2-1)
得：(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
=[(2+1)(2^2-1)(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+1)(2^128-1)+(2^2-1)]/(2^2-1)
=[(2+1)(2^128-1)]/(2^2-1)+1
=2^128
由于2^1=2 2^2=4 2^3=8 2^4=16
2^5=32 2^6=64
2^n末位是以(2,4,8,6)四个数为一周期循环
128/4=32能够整除，所以末位数是数字6