2^48-1可以被60和70之间某两个数整除,求这两个数,请给出详细过程,谢谢了
2^48-1=(2^24+1)(2^24-1)
=(2^24+1)(2^12+1)(2^12-1)
=(2^24+1)(2^12+1)(2^6+1)(2^6-1)
=(2^24+1)(2^12+1)*65*63
所以这两个数是65,63
=(2^24+1)(2^12+1)(2^12-1)
=(2^24+1)(2^12+1)(2^6+1)(2^6-1)
=(2^24+1)(2^12+1)*65*63
所以这两个数是65,63
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第1个回答 2008-03-23
2^48-1
=(2^24+1)(2^12+1)(2^6+1)(2^6-1)
=(2^24+1)(2^12+1)*63*65
所以这两个数是63和65
=(2^24+1)(2^12+1)(2^6+1)(2^6-1)
=(2^24+1)(2^12+1)*63*65
所以这两个数是63和65
第2个回答 2008-03-23
设2^6=t=64,则2^48-1=t^8-1=(t^4-1)*(t^4+1)
=(t^2-1)*(t^2+1)(t^4+1)
=(t-1)*(t+1)*(t^2+1)*(t^4+1)
所以这个数是t-1和t+1,即63和65
=(t^2-1)*(t^2+1)(t^4+1)
=(t-1)*(t+1)*(t^2+1)*(t^4+1)
所以这个数是t-1和t+1,即63和65
第3个回答 2008-03-23
我不是他舅 - 九门提督 终于在后面了!