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

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
aaronaa 1年前 已收到1个回答 举报

敏菜wm 幼苗

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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

1年前

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