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标题: 2005环球城市数学竞赛题 [打印本页]

作者: constant 时间: 2005-10-28 01:55
标题: 2005环球城市数学竞赛题

令N为正整数，已知方程式 99x+100y+101z = N 恰有一组正整数解，试求N的最大值。

作者: QL 时间: 2005-10-28 06:10
标题: 回复：2005环球城市数学竞赛题

5249?

作者: husonghu 时间: 2005-10-28 06:55
标题: [@};-]fzy may retire. constant has come. 懂了吗Jenny?

fzy may retire. constant has come. 懂了吗Jenny?

作者: ob 时间: 2005-10-28 08:07
标题: 回复：[@};-]fzy may retire. constant has come. 懂了吗Jen

you mean f change to c?

作者: husonghu 时间: 2005-10-28 08:26
标题: Very positive! [:D)][:D)][:D)]

Very positive!

作者: constant 时间: 2005-10-28 17:56
标题: Reason?

I don't know the exact number yet because I have not carried out all the calculation. It should be something in that range.

作者: constant 时间: 2005-10-28 18:51
标题: fzy 被 Jenny 和乱弹抓走了

fzy 被 Jenny 和乱弹抓走了

作者: Jenny 时间: 2005-10-28 18:56
标题: [;)]

作者: QL 时间: 2005-10-29 05:39
标题: 回复：Reason?

I guess the following idea should work:

Let N be the largest number and N = 99x+100y+101z,

since 2*100 = 99+101, we can see at once that

(1) y = 1 or 2

(2) min(x,z) = 1

Also,

99*51 = 101*49+100

101*50 = 99*50+100

Hence x<=51, z<=50

so(havn't check it carefully yet), the largest N is

99*1+100*2+101*50= 5349

作者: constant 时间: 2005-10-29 05:50
标题: 回复：回复：Reason?

5349 is a little too big. It is also 52*99 + 100 +101. I think the largest is 5251 = 50*99+2*100 + 101. But as usual I have not picked up a pen yet so not 100% sure.

作者: husonghu 时间: 2005-11-10 03:20
标题: 此题标准答案已由constant给出。有兴趣者请见坛上另贴。

此题标准答案已由constant给出。有兴趣者请见坛上另贴。

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