霹靂聊天會館

ariesgpgp

swordtrail大的問題好難呀~
我看了好幾遍才看懂它在問什麼
(太久沒算數學了= =)

呃，很抱歉
在下沒辦法幫到忙
數學實在不好啊 :(

[ 本帖最後由 ariesgpgp 於 2008-1-30 16:42 編輯 ]

玄字訣

大大們一樣的問候
依樣的時間
劣者來簽到啦
晚安

玄字訣

該睡覺囉
怎這幾天依然沒什麼人阿
呵呵不管啦先簽退囉

swordtrail

4. 找出最大整數 n, 使得 n 可整除所有小於 n 的立方根的正整數.

Don't worry...
It's a Olympiad Question, so it is meant to be hard...
Thanks for trying any way....

(Sigh, Sigh)

:P大家新年快樂!!!:victory:

[ 本帖最後由 swordtrail 於 2008-1-31 09:27 編輯 ]

swordtrail

原帖由 玄字訣 於 2008-1-31 01:16 發表
大大們一樣的問候
依樣的時間
劣者來簽到啦
晚安

I guess not much people are online, when you go...
M版主 has disappeared to for some reason too...
By the 玄字訣大, did you use to have an account called "giga"...

玄字訣

原帖由 swordtrail 於 2008-1-31 05:30 發表


I guess not much people are online, when you go...
M版主 has disappeared to for some reason too...
By the 玄字訣大, did you use to have an account called "giga"...

===============================================
有阿
可是自從三月浩劫讓我全部積分消失後
我就用新的也就是現在的 玄字訣了
感謝大大還記得我以前得戶名
好感動:')

玄字訣

大大們我來晚啦
不過還是要先簽到囉

玄字訣

QUOTE:
玄字訣大... just wondering did you use to have an account called "giga"...
Sorry, just wondering...
Because I think that account is still working...

===================================
劍大....好開心有人還記得我
有機會一起在這聊聊囉
有點晚了...該去睡囉
祝晚安
容劣者先行簽退

palpatine

  4. 找出最大整數 n, 使得 n 可整除所有小於 n 的立方根的正整數.

這題目出錯了吧，n怎麼可能整除比它小的數呢，除非這個比它小的數是零。

swordtrail

原帖由 玄字訣 於 2008-2-1 03:01 發表
QUOTE:
玄字訣大... just wondering did you use to have an account called "giga"...
Sorry, just wondering...
Because I think that account is still working...

========================= ...

giga (嵐刀)
中級會員
UID 123572
精華 0
積分 170
帖子 67
閱讀權限 30
註冊 2006-5-27

I remember I dropped from 中級會員 to 註冊會員 after 三月浩劫...
(But at least I made it back...)
Very sorry...
School is going to start soon...
My time on the computer is rather limited...
I'll be happy to chat when I am free...
(But you need to come earlier....
NZ time = Taiwan time + 5 hrs...)



:P大家新年快樂!!!:victory:

[ 本帖最後由 swordtrail 於 2008-2-1 09:39 編輯 ]

swordtrail

原帖由 palpatine 於 2008-2-1 09:21 發表


這題目出錯了吧，n怎麼可能整除比它小的數呢，除非這個比它小的數是零。

I think you may have misunderstood the question...
It means that every positive integer less than the cube root of n is divisable of n...
Does it make it clearer that way???
Can you help???

palpatine

Well, I think you translated the problem incorrectly.
However, I got your idea. I can give you a hint. We have two steps to go.

1st, you have to give an upper bound of the candidates. Second, you test all the candidates in a smart way. Then you can figure out which one is the greatest number satisfying this divisible property.

The second step is an easy routine. To achieve the first step, we denote the candidate number as M=A*b.
Suppose  n < =  M^(1/3) < n+1. Here we assume A is the least common multiple of 1,2,3,....,n, and   b is some positive integer.  Therefore, we get
n^3  <=  A*b  < (n+1)^3 .
We want to use the right hand side to get an upper bound of A. Namely,
  A < = A*b < (n+1)^3
Once you derive an upper bound of M from the above inequality, you can easily get the answer. It takes efforts to conquer this first step. Hope this helps. Good luck! Young man!

   



   

原帖由 swordtrail 於 2008-2-1 05:29 發表


I think you may have misunderstood the question...
It means that every positive integer less than the cube root of n is divisable of n...
Does it make it clearer that way???
Can you help???

palpatine

原帖由 palpatine 於 2008-2-1 05:49 發表
Well, I think you translated the problem incorrectly.
However, I got your idea. I can give you a hint. We have two steps to go.

1st, you have to give an upper bound of the candidates. Second,  ...

An lower bound of A might help you more:

It is easy to see that n and n-1 are coprime. Therefore A>=n(n-1)
Then we check n-2. n-2 and n-1 are coprime.
1> If n is an odd number, n-2 and n are coprime. This implies A > = n(n-1)(n-2).
     Next, check n-3. n-3 is an even number (since n is odd). Therefore (n-3)/2  and n-1 are coprime. The only possible common divisor of n-3 and n is 3. (Note that n/3 - (n-3)/3 =1). This implies A>=n(n-1)(n-2)(n-3)/6
2> If n is even, follow the same idea as I show in the previous step. You can still conclude A>= n(n-1)(n-2)(n-3)/6

With this lower bound of A, you can easily solve the upper bound of n and move to the routine step. Then you are almost there!

swordtrail

原帖由 palpatine 於 2008-2-1 09:49 發表
Well, I think you translated the problem incorrectly.
However, I got your idea. I can give you a hint. We have two steps to go.

1st, you have to give an upper bound of the candidates. Second,  ...

Thank you for your hint...
But there is one part I don't quite understand...

A < = A*b < (n+1)^3

Since A*b = M, and (n+1) > M^(1/3) Then isn't it given that M < (n+1)^3???

palpatine

原帖由 swordtrail 於 2008-2-1 06:37 發表


Thank you for your hint...
But there is one part I don't quite understand...

Since A*b = M, and (n+1) > M^(1/3) Then isn't it given that M < (n+1)^3???

Yes. Actually, we derive A<= (n+1)^3 from that fact M<(n+1)^3. The useful part is A < = (n+1)^3. As you can see in the further hint, I derive a lower bound of A in terms n. Then we get an inequality in terms of n. Hence we can estimate the upper bound of n.

swordtrail

原帖由 palpatine 於 2008-2-1 10:42 發表


Yes. Actually, we derive A<= (n+1)^3 from that fact M<(n+1)^3. The useful part is A < = (n+1)^3. As you can see in the further hint, I derive a lower bound of A in terms n. Then we get ...

I need some time to digest this...
I'll think about it...
Thank you very much for your help...
Just wondering, do you have any background in Mathematics???

palpatine

I would say the only background in my life is Mathematics. You are on the trail I have  passed long time ago. Good luck!

原帖由 swordtrail 於 2008-2-1 06:45 發表


I need some time to digest this...
I'll think about it...
Thank you very much for your help...
Just wondering, do you have any background in Mathematics???

onep

原帖由 swordtrail 於 2008-2-1 05:29 發表
I think you may have misunderstood the question...
It means that every positive integer less than the cube root of n is divisable of n...

看起來
正確的中譯應該是
找出最大整數 n, 使得 n 可以被所有小於 n 的立方根的正整數整除