Gopalgang higher hisec 2011 08
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S={1,2,3,.........,441}&X is a subset of S that no two member of X is a square less or equal to 21^2.What is the largest number of elements can X have?
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Re: Gopalgang higher hisec 2011 08
this may go thus:
in $S$ there are $21$ squares.which all are less or equal to $21^2$ . again, we can have a square number like that. so the answers is $441-21+1=420$ which is the highest number of elements of $X$
in $S$ there are $21$ squares.which all are less or equal to $21^2$ . again, we can have a square number like that. so the answers is $441-21+1=420$ which is the highest number of elements of $X$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Gopalgang higher hisec 2011 08
This solution is not correct...
eg 2+7=9... A SQUARE!!
The solution might be 221... But need a full solution!!
eg 2+7=9... A SQUARE!!
The solution might be 221... But need a full solution!!
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Re: Gopalgang higher hisec 2011 08
vaia, i could not get where the problem is because the problem says "X is a subset of S that no two member of X is a square less or equal to 21^2"
the meaning of this i got is such "there is only one number in X which is less or equal to $21^2$ at most"
as it doesn't say 'summation or total of two numbers less or equal to $21^2$'
the meaning of this i got is such "there is only one number in X which is less or equal to $21^2$ at most"
as it doesn't say 'summation or total of two numbers less or equal to $21^2$'
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Gopalgang higher hisec 2011 08
আচ্ছা এখানে X এর ২টি সদস্যকে কি যোগ করতে বলেছে? যদি না বলে তাহলে নাফিসেরটা সঠিক টা না হলে আরো ঝামেলা আচে।Labib wrote:This solution is not correct...
eg 2+7=9... A SQUARE!!
The solution might be 221... But need a full solution!!
অম্লান সাহা
Re: Gopalgang higher hisec 2011 08
শিশির ভাই এর কম্পোজিং ভুল!
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
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Re: Gopalgang higher hisec 2011 08
আগে বলবেন না ! এখন তো অনেক জটিল হয়ে গেল ।
এখন, প্রতিটা বর্গ সংখ্যার partition বের করে সেগুলো বাদ দিতে হবে ।
অর্থাৎ,
$x^2=y+z$ হলে $y$ আর $z$ একসাথে থাকা যাবে না । আমি brute force ছাড়া তো পথ দেখতেছি না।
এখন, প্রতিটা বর্গ সংখ্যার partition বের করে সেগুলো বাদ দিতে হবে ।
অর্থাৎ,
$x^2=y+z$ হলে $y$ আর $z$ একসাথে থাকা যাবে না । আমি brute force ছাড়া তো পথ দেখতেছি না।
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Gopalgang higher hisec 2011 08
may be this is the solution :
as it is said that summation of no two numbers can be $\leq 21^2$ and the largest member of $S$ id $441$
so the answer will be
\[\left \lfloor \frac{21^2}{2} \right \rfloor+1=221\]
i think there is some incompleteness of the solution .but, probably the way is right.
the incompleteness is we can add another number.which is $441$
so, the answer is
\[222\]
as it is said that summation of no two numbers can be $\leq 21^2$ and the largest member of $S$ id $441$
so the answer will be
\[\left \lfloor \frac{21^2}{2} \right \rfloor+1=221\]
i think there is some incompleteness of the solution .but, probably the way is right.
the incompleteness is we can add another number.which is $441$
so, the answer is
\[222\]
Last edited by nafistiham on Tue Dec 06, 2011 12:46 am, edited 1 time in total.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Gopalgang higher hisec 2011 08
No compose was right . I have copied it from the original question
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Re: Gopalgang higher hisec 2011 08
তাহলে তো সমাধানটা নিশ্চয়ই ওভাবেই হবে । তাই না ?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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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