Question

试确定(2+1)(2^2+1)(2^4+1)2^8+1)(2^16+1)(2^32+1)(2^64+1)+1的末尾数字

RT 正确答案是6 求过程、

Answer 1

(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
可以先求出前面乘积的未尾数字
2 2
2*2 4
2^4 6
2^8 6
2^16 6
2^32 6
2^64 6
加一之后乘起来
7^5*3*5
7 7
7^2 9
7^3 3
7^4 1
7^5 7
7*3*5+1
最后一位是
6

Answer 2

(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
=(2^16-1)(2^16+1)(2^32+1)(2^64+1)+1
=(2^32-1)(2^32+1)(2^64+1)+1
=(2^64-1)(2^64+1)+1
=2^128-1+1
=2^128
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
...
根据以上规律可知每4次方后，尾数重复
128/4=32
所以2^128的尾数是6