APMO 1999-2
What is the largest integer divisible by all positive integers less than its cube root?
One one thing is neutral in the universe, that is $0$.
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Re: APMO 1999-2
Answer=5. $n!$ divisor of $n^3$. sterlimg's formula\[n!~\sqrt{2\left ( pi \right )n}\left ( n/e \right )^{n}\] then \[n>\left ( n/e \right )^{n}\]
\[n>\left ( n/e \right )^{\left ( n-2 \right )}\]
\[e^{n-2/n-3}>n\]
\[n>\left ( n/e \right )^{\left ( n-2 \right )}\]
\[e^{n-2/n-3}>n\]
Re: APMO 1999-2
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.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 \[n>\left ( n/e \right )^{n}\]
\[n>\left ( n/e \right )^{\left ( n-2 \right )}\]
\[e^{n-2/n-3}>n\]
One one thing is neutral in the universe, that is $0$.
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- Posts:56
- Joined:Fri Feb 18, 2011 11:30 pm
Re: APMO 1999-2
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 ?
Re: APMO 1999-2
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.
One one thing is neutral in the universe, that is $0$.
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- Posts:56
- Joined:Fri Feb 18, 2011 11:30 pm
Re: APMO 1999-2
cube root of 420 not an ineger is not it a problem ?
Re: APMO 1999-2
No, it meant \[[1,2,\ldots\ldots,\lfloor \sqrt[3]{n}\rfloor]|n\]
One one thing is neutral in the universe, that is $0$.