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.

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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)(2^128+1),求M的个位数字.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)(2^128+1),求M的个位数字.
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.

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.
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

因为只要求M的个位数字
所以只需看各个因子的个位数
3*5*7*7*7*7*7*7=……5
所以个位数字为5