(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.

来源：学生作业帮助网编辑：作业帮时间：2024/05/11 16:20:05

(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.
(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.

(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.
(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.

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(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

收起

1/2,1/4, (1-1/2^2)(1-1/3^2)(1-1/4^2).(1-1/2009^2), (2+1)({2}^{2}+1)({2}^{4}+1)({2}^{8}+1).({2}^{64}+1)+1 巧算((2^1+1)(2^2+1)(2^4+1)(2^8+1)+1)/2^15 计算(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1 计算:(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16.计算：(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16. [(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(1+2^-(1/2)] (1-1/2^2)*(1-1/3^2)*(1-1/4^2)*.*(1-1/2002^2)*(1-1/2003^2) (1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16), 1,2,4,4,1,( ) (2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1 (2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1） (2+1)(2-1)(2^2+1)(2^4+1).(2^8+1)化简 (1+2)*(1+2^2)*(1+2^4)*(1+2^8)*(1+2^16) 化简(1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16) 化简(2+1)(2^2+1)(2^4+1)(2^8+1)…(2^256+1) (2-1)(2+1)(2^2+1)(2^4)...(2^64+1)+1=? (2+1)(2^2+1)(2^4+1)(2^8+1)……(2^1024+1)