作业帮 > 数学 > 作业

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

来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/05/29 12:01:16
已知M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1) 求M个位数是几
已知M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128
M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^16-1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^32-1)(2^32+1)(2^64+1)(2^128+1)
=(2^64-1)(2^64+1)(2^128+1)
=(2^128-1)(2^128+1)
=2^256 - 1
个位数 与2^4-1的个位数相同,为5