Question

数学题 因式分解

已知 2^96 一 1 可被在60~70之间的两个整数整除，求这两个整数。（写出详细过程）

Answer 1

解：2^96-1
=(2^48+1)(2^48-1)
=(2^48+1)(2^24+1)(2^24-1)
=(2^48+1)(2^24+1)(2^12+1)(2^12-1)
=(2^48+1)(2^24+1)(2^12+1)(2^6+1)(2^6-1)
因为2^6=64
所以2^6+1=65,2^6-1=63
答:这两个数是63和65.

Answer 2

解：2^96-1
=(2^48+1)(2^48-1)
=(2^48+1)(2^24+1)(2^24-1)
=(2^48+1)(2^24+1)(2^12+1)(2^12-1)
=(2^48+1)(2^24+1)(2^12+1)(2^6+1)(2^6-1)
=(2^48+1)(2^24+1)(2^12+1)×65×63
这两个数是63和65

Answer 3

解：在给2^96 一 1分解因式时
必有(2^6+1)和（2^6-1)这两个因式
所以这两个整数是(2^6+1）=64+1=65
（2^6-1)=64-1=63
所以这两个整数是63和65

Answer 4

2^96 一 1 =(2^48+1)(2^48-1) = (2^48+1)(2^24+1) (2^24-1)= ……= (2^48+1)(2^24+1) (2^12+1)（2^6+1)(2^3+1)(2^3-1)= (2^48+1)(2^24+1) (2^12+1)*65* 9*7=(2^48+1)(2^24+1) (2^12+1)*65* 63,
所以2^96 一 1 可被在60~70之间的两个整数整除的两个整数分别为65和63

Answer 5

2^96-1=(2^48)^2-1^2=(2^48+1)(2^48-1)=(2^48+1)(2^24+1)(2^24-1)=(2^48+1)(2^24+1)(2^12+1)(2^12-1)=(2^48+1)(2^24+1)(2^12+1)(2^6+1)(2^6-1)=65*67(2^48+1)(2^24+1)(2^12+1)
这两个数是65和67

Answer 6

1.利用平方差公式计算，有
2^96 一 1
=(2^48+1)(2^48-1)
=(2^48+1)(2^24+1)(2^24-1)
=(2^48+1)(2^24+1)(2^12+1)(2^12-1)
=(2^48+1)(2^24+1)(2^12+1)(2^6+1)(2^6-1)
因为2^6-1=64-1=63，
2^6+1=65
所以可以被63和65整除

Answer 7

2^96-1=(2^48+1)(2^48-1)=(2^48+1)(2^24+1)(2^24-1)=(2^48+1)(2^24+1)(2^12+1)(2^12-1)
=(2^48+1)(2^24+1)(2^12+1)(2^6+1)(2^6-1)
因为2^6+1=65 2^6-1=63
所以在2^96 一 1 可被在60~70之间的两个整数整除的两个整数是65和63