Question

2的128次方加2的和除以3 要过程

Answer 1

(2^128+2)/3
=(2^128-1+3)/3
=[(2^128-1)/3]+1
=(2^64+1)(2^64-1)+1
=......................................
=(2^64+1)(2^32+1)(2^16+1)(2^8+1)(2^4+1)(2^2+1)(2^2-1)+1
=3*(2^64+1)(2^32+1)(2^16+1)(2^8+1)(2^4+1)(2^2+1)+1

所以：
最终结果(2^128+2)/3=(2^64+1)(2^32+1)(2^16+1)(2^8+1)(2^4+1)(2^2+1)+1

Answer 2

(2^128+2)-3
=2^128-1
=(2^64+1)(2^64-1)
=(2^64+1)(2^32+1)(2^32-1)
=......
=(2^64+1)(2^32+1)(2^16+1)(2^8+1)(2^4+1)(2^2+1)(2^2-1)
=3*(2^64+1)(2^32+1)(2^16+1)(2^8+1)(2^4+1)(2^2+1)

所以：
(2^128+2)/3=1+(2^64+1)(2^32+1)(2^16+1)(2^8+1)(2^4+1)(2^2+1)

追问

是除以3不是减3 但还是谢谢你

追答

请看清楚，最后的结果就是除以3的，(2^128+2)/3