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

问题描述:

设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的个位数字.
1个回答 分类:数学 2014-09-26

问题解答:

我来补答
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
2的1次方个位数是:2
2的2次方个位数是:4
2的3次方个位数是:8
2的4次方个位数是:6
2的5次方个位数是:2
∴2的n次方个位数是以:2、4、8、6四个数为循环的
∵256÷4=64
∴2^256的个位数是:6
∴2^256-1的个位数是:5
即:M的个位数是:5
再问: 请问能告诉我要是写过程要怎么写吗?
再答: 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) …… =2^256-1 ∵2的n次方个位数是以:2、4、8、6四个数为循环的,256÷4=64 ∴2^256的个位数是:6 ∴2^256-1的个位数是:5 这样写就行了!
 
 
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