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[数论] 转发QQ群看到一题

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realnumber Posted 2015-6-22 22:13 |Read mode
Last edited by realnumber 2015-6-22 22:21严: QQ图片20150622220936.png

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 Author| realnumber Posted 2015-6-22 22:18
QQ图片20150622221202.png

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&活着存在 Posted 2015-6-24 11:07
Last edited by &活着存在 2015-6-24 11:13 未命名.JPG

原来那号"活着&存在"怎么不能登录?
是系统故障,还是号已经中毒?

未命名.JPG

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kuing Posted 2015-6-24 12:17
回复 3# &活着存在

系统没问题,请确认帐号及密码是否正确输入,比如说大小写、全角半角有没有切换错等

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 Author| realnumber Posted 2015-6-24 15:06
回复 3# &活着存在
en,一楼的问题还是没解决,
不妨设$1\le a_1<a_2<a_3<\cdots <a_{n},n>3$,这n个正整数符合“所有数的和被任意2数和整除”.
若其公因数为$m=(a_1,a_2,\cdots,a_{n})$,
则$\frac{a_1}{m},\frac{a_2}{m},\cdots,\frac{a_{n}}{m}$也符合要求.
因此不妨设m=1.
当n=3时,$a_1+a_2\le a_3,a_2+a_3\le a_1$矛盾.
当n=4时,方法同n=3.
继续想.....

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Tesla35 Posted 2017-1-10 22:10
都是镇中特长生招生试题。

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