Advanced P-3(BOMC-2)

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Phlembac Adib Hasan
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Re: Advanced P-3(BOMC-2)

Unread post by Phlembac Adib Hasan » Sun Apr 01, 2012 12:50 pm

I have come from school only a few minutes ago and solved this problem first.
Solution :
For odd numbers>1 just take $k$ and $k+1$ to get $2k+1$ for $k\in \mathbb {N}$.So the even case is remaining.
If we sum up $2k+1$ consecutive numbers then the result is $(m-k)+(m-(k-1))+....+m+m+1+...+(m+k)=(2k+1)m$
which can handle any even number excluding the powers of two.
Now notice if sum up $2k$ consecutive numbers then the result is $(m-k)+(m-(k-1))+....+m+m+1+...+(m+k-1)=2mk-k=k(2m-1)$ which can't be a power of $2$ as $2m-1$ is always odd.
So the powers of two are the number those can't be expressed as sum of consecutive integers.
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Re: Advanced P-3(BOMC-2)

Unread post by *Mahi* » Sun Apr 01, 2012 1:07 pm

Phlembac Adib Hasan wrote:I have come from school only a few minutes ago and solved this problem first.
Solution :
For odd numbers>1 just take $k$ and $k+1$ to get $2k+1$ for $k\in \mathbb {N}$.So the even case is remaining.
If we sum up $2k+1$ consecutive numbers then the result is $(m-k)+(m-(k-1))+....+m+m+1+...+(m+k)=(2k+1)m$
which can handle any even number excluding the powers of two.
Now notice if sum up $2k$ consecutive numbers then the result is $(m-k)+(m-(k-1))+....+m+m+1+...+(m+k-1)=2mk-k=k(2m-1)$ which can't be a power of $2$ as $2m-1$ is always odd.
So the powers of two are the number those can't be expressed as sum of consecutive integers.
This is only half the case.What if $k=2^x>m$?
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Re: Advanced P-3(BOMC-2)

Unread post by Phlembac Adib Hasan » Sun Apr 01, 2012 3:16 pm

I am not realizing.Will you please explain?I have showed for every odd number>1 and for all the even numbers other than powers of two we can express them as sum of some consecutive numbers.Will you please give an example which part I am missing?
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Re: Advanced P-3(BOMC-2)

Unread post by *Mahi* » Sun Apr 01, 2012 3:18 pm

sourav das wrote:Most positive integers can be expressed as a sum of two or more consecutive
positive integers. For example, $24 = 7 + 8 + 9$ and $51 = 25 + 26$. A
positive integer that cannot be expressed as a sum of two or more consecutive
positive integers is therefore interesting. What are all the interesting
integers?
The red part.
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Re: Advanced P-3(BOMC-2)

Unread post by Niloy Da Fermat » Sun Apr 01, 2012 3:28 pm

i still don't get it.we are selecting $ m $ st $ m $ is always greater than $ k $.isn't it?
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Re: Advanced P-3(BOMC-2)

Unread post by *Mahi* » Sun Apr 01, 2012 3:31 pm

Niloy Da Fermat wrote:i still don't get it.we are selecting $ m $ st $ m $ is always greater than $ k $.isn't it?
Then try choosing $m,k$ from $24=3 \times 8 = 3 \cdot 2^3 =(2 \cdot 1+1) 2^3 = (2 \cdot 2 -1) 2^3$
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Re: Advanced P-3(BOMC-2)

Unread post by Phlembac Adib Hasan » Sun Apr 01, 2012 4:15 pm

Why?Here I'll take $m=8$,$k=1$.I am still not realizing.
I said I can find some $k$ and $m$ for every even number except the powers of two.,But I am not giving any exact value.I think you are pointing to this:How can it be true for $10$?It can't be expressed as sum of five consecutive integers.But notice that I also wrote something about the even case.
Here $10=2.5=2.(2.3-1)$.Hence we have to take four numbers.So I can take $m>k$.
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Re: Advanced P-3(BOMC-2)

Unread post by Ehsan » Sun Apr 01, 2012 4:44 pm

It's the problem 2.4.15 from The Art And Craft of Problem Solving.
There it was said, "algebra can be used, but so can pictures."
Try solving by picture, it's much interesting and simple.
Every even integer is a sum of two primes.

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Re: Advanced P-3(BOMC-2)

Unread post by Masum » Sun Apr 01, 2012 5:18 pm

You can actually conclude that if a number can be expressed as a sum of some consecutive numbers in $d(n)$ ways, then $d(n)$ is the number of odd divisors of $n$. That also makes a sense why powers of $2$ won't satisfy the condition.(reverse thinking)
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Re: Advanced P-3(BOMC-2)

Unread post by Phlembac Adib Hasan » Tue Apr 03, 2012 12:51 pm

Where's my mistake?Anyone please explain.
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