设M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),试确定M-2006的个位数字是什么?
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设M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),试确定M-2006的个位数字是什么?设M=(2+1)(2^2+1)(2^4+1)(2^8+1
设M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),试确定M-2006的个位数字是什么?
设M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),试确定M-2006的个位数字是什么?
设M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),试确定M-2006的个位数字是什么?
M=(2+1)(2^2+1)(2^4+1)...(2^64+1)=[(2-1)(2+1)](2^2+1)(2^4+1)...(2^64+1)=[(2^4-1)(2^4+1)](2^8+1)(2^16+1)...(2^64+1)=...=2^128-1,因2的几次方的个位数分别是:2,4,8,6,2,4...4,又128/4=32,所以2^128的个位数与2^4的个位数相同,为6;所以的M个位数是6-1=5.
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