APMO 1999-2

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Masum
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APMO 1999-2

Unread post by Masum » Sat Mar 05, 2011 1:54 pm

What is the largest integer divisible by all positive integers less than its cube root?
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tarek like math
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Re: APMO 1999-2

Unread post by tarek like math » Fri Apr 29, 2011 5:05 pm

Answer=5. $n!$ divisor of $n^3$. sterlimg's formula\[n!~\sqrt{2\left ( pi \right )n}\left ( n/e \right )^{n}\] then \[n>\left ( n/e \right )^{n}\]
\[n>\left ( n/e \right )^{\left ( n-2 \right )}\]
\[e^{n-2/n-3}>n\]

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Re: APMO 1999-2

Unread post by Masum » Sat Apr 30, 2011 2:56 pm

tarek like math wrote:Answer=5. $n!$ divisor of $n^3$. sterlimg's formula\[n!~\sqrt{2\left ( pi \right )n}\left ( n/e \right )^{n}\] then \[n>\left ( n/e \right )^{n}\]
\[n>\left ( n/e \right )^{\left ( n-2 \right )}\]
\[e^{n-2/n-3}>n\]
What about $n=420$ then? In fact such type of strict formulas can hardly help us to solve a contest problem. Try with your intution which can really help you. Also use a correct spelling. If needed, search in Wikepedia.
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Re: APMO 1999-2

Unread post by tarek like math » Sun May 01, 2011 1:21 am

the thing i understand is find a largest integer suppose $n^3$ which divisible by all integer less than n means all 1~n numbers are factor of $n^3$. then n=420 or $n^3$=420 not correct value. what u mean ?

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Re: APMO 1999-2

Unread post by Masum » Tue May 03, 2011 2:21 pm

First understand a problem what it wants really. Then take your approach. You didn't understand what this problem meant. There is enough language to understand.
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Re: APMO 1999-2

Unread post by tarek like math » Tue May 03, 2011 2:59 pm

cube root of 420 not an ineger is not it a problem ?

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Re: APMO 1999-2

Unread post by Masum » Wed May 11, 2011 2:20 pm

No, it meant \[[1,2,\ldots\ldots,\lfloor \sqrt[3]{n}\rfloor]|n\]
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