若n^3+100可以被n+10所整除,求n的最大整數是多少?
回答 (11)
2014-04-16 6:38 pm
✔ 最佳答案
若n^3+100可以被n+10所整除,求n的最大整數是多少?Sol
(n^3+100)/(n+10)=n^2－10n+100－900/(n+10)
(n+10)｜900
n的最大整數=890
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2014-04-16 6:44 pm
n+10 l n^3+100
==>n+10 l (n^3+100)-n^2(n+10)
==>n+10 l 100-10n^2
==>n+10 l (100-10n^2)+10n(n+10)
==>n+10 l 100(1+n)
==>100(1+n)=(n+10)*p , p不等於0
因1+n不等於n+10 ,所以100=n+10 ==>n=90
==>n+10 l (n^3+100)-n^2(n+10)
==>n+10 l 100-10n^2
==>n+10 l (100-10n^2)+10n(n+10)
==>n+10 l 100(1+n)
==>100(1+n)=(n+10)*p , p不等於0
因1+n不等於n+10 ,所以100=n+10 ==>n=90
收錄日期: 2021-04-30 18:44:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140416000015KK01903