from 1 to 100
take any two number from 1 to 100.add them.how many results will be same
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nafistiham
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Re: from 1 to 100
$1+4=5$
$2+3=5$
$2+4=6$
$1+5=6$
there are many like these. What is the want of the problem. Is $same$ $sum$ means, no matter how, if two pairs get the same summation ?
Or, are you talking about the highest number of pairs, which will have the same sum ?
If the first assumption is right. only, $1+2=3$ and $100+99=199$ can be done in just one way. All the sums other pairs will have a twin.
But, I believe, that is not it. So, kindly, clarify.
$2+3=5$
$2+4=6$
$1+5=6$
there are many like these. What is the want of the problem. Is $same$ $sum$ means, no matter how, if two pairs get the same summation ?
Or, are you talking about the highest number of pairs, which will have the same sum ?
If the first assumption is right. only, $1+2=3$ and $100+99=199$ can be done in just one way. All the sums other pairs will have a twin.
But, I believe, that is not it. So, kindly, clarify.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Phlembac Adib Hasan
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Re: from 1 to 100
I believe in the at most case, too.
This question is related to $P(n,2)$. For $n>101$, there are exactly $P(n,2)-(n-101)$ pairs satisfying the condition. And for $n\leq 101$, $P(n,2)$ is the answer. Here $P(n,2)\leq P(101,2)=50$
Now notice that \[\left \lfloor \frac{n}{2} \right \rfloor -n+101<\frac{n}{2}-n+101=101-\frac{n}{2}<101-\frac{102}{2}=50\]
Therefore $P(101,2)$ is (one of) the highest possible value(s) and there will be at most $50$ pairs.
This question is related to $P(n,2)$. For $n>101$, there are exactly $P(n,2)-(n-101)$ pairs satisfying the condition. And for $n\leq 101$, $P(n,2)$ is the answer. Here $P(n,2)\leq P(101,2)=50$
Now notice that \[\left \lfloor \frac{n}{2} \right \rfloor -n+101<\frac{n}{2}-n+101=101-\frac{n}{2}<101-\frac{102}{2}=50\]
Therefore $P(101,2)$ is (one of) the highest possible value(s) and there will be at most $50$ pairs.
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Re: from 1 to 100
i want the answer that how many answers will be same.for example 100 can be got in 49 ways,99 can be 50 ways....i want the total sum.
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nafistiham
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Re: from 1 to 100
Let us see how much far I can goRafe wrote:i want the answer that how many answers will be same.for example $100$ can be got in $49$ ways, $99$ can be $50$ ways....i want the total sum.
we can get, $5,6...197$ as same sums. $5,6;7,8;9,10;11,12;.........;99,100$ can be got in $2;3;4;5.......;49$ ways.
now, we have to go $101$ to $197$
suppose, taking numbers from $1$ to $100$ is not compulsory.
So, $101,102;103,104;105,106;107.........;195,196;197,198$ can be got in $50;51;52;......;98$ ways.
But, we can't take $n>100$ so, we have to delete the cases like $104=101+3=102+2;105=101+4=102+3=103+2=104+1$
like these for $102;103;104;105;.......;197;198$, the number of deleted cases will be $1;2;3;4;.....;96;97$ ways.
So, now, let us sum up.
the total number will be
\[48 \cdot 51 + 148 \cdot 49 - \frac {97 \cdot 98} {2} - 1\]
The red part is for $99$ can be done in $49$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: from 1 to 100
there is a very shortcut solution.can anyone find it?
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nafistiham
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Re: from 1 to 100
Share it, kindly, then.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: from 1 to 100
We can take $\displaystyle \binom {100}{2}=4950$ pairs. Now the least result is $1+2=3$ and the greatest is $100+99=199$. So only $197$ results are different and $4950-197$ results are same.
Last edited by Phlembac Adib Hasan on Sat Jan 19, 2013 7:20 pm, edited 1 time in total.
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nafistiham
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Re: from 1 to 100
That was easy !!Rafe wrote:We can take $\displaystyle \binom {100}{2}=4950$ pairs. Now the least result is $1+2=3$ and the greatest is $100+99=199$. So only $197$ results are different and $4950-197$ results are same.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.