a=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^61+1),则a-1996的应该是:_数学解答

a=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^61+1),则a-1996的
应该是:a=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),则a-1996的末位数字....

人气：383 ℃　时间：2020-04-15 15:24:05

解答

给a乘以一个1,a不变,而1＝2－1,故可以连续使用平方差公式.a＝(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)＝(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)＝(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^...