Kustia 2009 HS prob 7
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$2, 7, 11, 13, 69, 111, 420$ সংখ্যাগুলোকে কতভাবে সাজানো যায় যেন পাশাপাশি যেকোনো চারটি সঙ্খ্যার সমষ্টি সব সময় $3$ দ্বারা বিভাজ্য হয়?
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Re: Kustia 2009 HS prob 7
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Nur Muhammad Shafiullah | Mahi
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nafistiham
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Re: Kustia 2009 HS prob 7
there are $3$ types of number here.suppose,
\[3n+1=X\]
\[3n=Z\]
\[3n-1=Y\]
the condition is satisfied when it is
\[XZYZXZY\]
\[or\]
\[YZXZYZX\]
the number of $X,Y,Z$ are such $2,2,3$
so the number of combinations are
\[2\cdot2\cdot3!=24\]
it is not correct.
please see the next posts for futher correction.
\[3n+1=X\]
\[3n=Z\]
\[3n-1=Y\]
the condition is satisfied when it is
\[XZYZXZY\]
\[or\]
\[YZXZYZX\]
the number of $X,Y,Z$ are such $2,2,3$
so the number of combinations are
\[2\cdot2\cdot3!=24\]
it is not correct.
please see the next posts for futher correction.
Last edited by nafistiham on Tue Dec 06, 2011 1:27 pm, edited 2 times in total.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Kustia 2009 HS prob 7
nafis, as u have considered two situations(xzyzxzy & yzxzyzx) so you have to multiply $24$ with another $2$. so answer should be 48nafistiham wrote: \[XZYZXZY\]
\[or\]
\[YZXZYZX\]
the number of $X,Y,Z$ are such $2,2,3$
so the number of combinations are
\[2\cdot2\cdot3!=24\]
অম্লান সাহা
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nafistiham
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Re: Kustia 2009 HS prob 7
yap.thanks.
missed that.
missed that.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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Re: Kustia 2009 HS prob 7
@মাহি, আমি Mod 3 নিয়াই চিন্তা করসি।
@অম্লান দা + তিহাম- আমি ঠিকভাবে বের করতে পারসি... (Courtesy - Bristy)
At first I also thought that there were only 48.
But look at these combos::
$ZXYZZXY$ and $XYZZXYZ$.
In the same way they have 48 permutations.
So the solution is $144$.
@অম্লান দা + তিহাম- আমি ঠিকভাবে বের করতে পারসি... (Courtesy - Bristy)
At first I also thought that there were only 48.
But look at these combos::
$ZXYZZXY$ and $XYZZXYZ$.
In the same way they have 48 permutations.
So the solution is $144$.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Re: Kustia 2009 HS prob 7
hmm..... so there are total 3 combos, each having 48 permutationsLabib wrote: At first I also thought that there were only 48.
But look at these combos::
$ZXYZZXY$ and $XYZZXYZ$.
In the same way they have 48 permutations.
So the solution is $144$.
$XZYZXZY$
$ZYXZZYX$
$ZXYZZXT$
and the answer is 144
thanks Labib
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