Question

求个位数字 求（2-1)(2+1)(2^2+1)(2^4+1)...(2^32+1)+1的个位数字 用代数式说明

Answer 1

原式=3(2^2+1)(2^2-1)(2^4+1)(2^4-1)...(2^32+1)(2^32-1)/(2^2-1)(2^4-1)...(2^32-1)+1
=3(2^4-1)(2^8-1)...(2^64-1)/(2^2-1)(2^4-1)...(2^32-1)+1
=2^64-1+1
=2^64
2^16 = 65536
所以（2-1)(2+1)(2^2+1)(2^4+1)...(2^32+1)+1的个位数字也为6
第2种解法
（2-1)(2+1)(2^2+1)(2^4+1)...(2^32+1)+1=2^64,
2^4=16,2^64是16个2^4相乘,6乘6的尾数恒为6.所以
（2-1)(2+1)(2^2+1)(2^4+1)...(2^32+1)+1个位数是6