Multiple of 5

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Fahim Shahriar
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Multiple of 5

Unread post by Fahim Shahriar » Sun Dec 30, 2012 12:58 am

How many $8$ digit numbers are there whose sum of digits; is a multiple of $5$?
Name: Fahim Shahriar Shakkhor
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Phlembac Adib Hasan
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Re: Multiple of 5

Unread post by Phlembac Adib Hasan » Sun Dec 30, 2012 6:24 pm

It is zero included partition.
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SANZEED
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Re: Multiple of 5

Unread post by SANZEED » Sun Dec 30, 2012 10:13 pm

Phlembac Adib Hasan wrote:It is zero included partition.
Well,you need to take in account the permutations too.
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nafistiham
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Re: Multiple of 5

Unread post by nafistiham » Mon Dec 31, 2012 9:31 pm

What if we start like this,
$10000004$ is the first one.
$10000009$ is the next.
$10000013,10000018,10000022,10000027$
don't we get $2$ in every $10$ ?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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*Mahi*
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Re: Multiple of 5

Unread post by *Mahi* » Tue Jan 01, 2013 12:56 am

nafistiham wrote: don't we get $2$ in every $10$ ?
This observation is so true that it kills the problem in just one line :D
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nafistiham
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Re: Multiple of 5

Unread post by nafistiham » Sat Jan 05, 2013 9:15 pm

*Mahi* wrote:
nafistiham wrote: don't we get $2$ in every $10$ ?
This observation is so true that it kills the problem in just one line :D
The truth is that, I posted this 'post' as an observation because I could not find out a rigorous 'chopping' logic after a lot of brain cooking.
I believe, someone will do that.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Multiple of 5

Unread post by kfoozminus » Sun Jan 06, 2013 2:08 pm

$\frac{99999999-9999999}{10}\cdot2=18000000$

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Re: Multiple of 5

Unread post by *Mahi* » Sun Jan 06, 2013 11:36 pm

nafistiham wrote: The truth is that, I posted this 'post' as an observation because I could not find out a rigorous 'chopping' logic after a lot of brain cooking.
I believe, someone will do that.
Proof:
Let the sum of first 7 digits of the number be $k$. then there will be exactly $2$ numbers between $0-9$ which, when added with $k$, is divisible by $5$.
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Re: Multiple of 5

Unread post by kfoozminus » Mon Jan 07, 2013 1:07 pm

simple logic: starting with $10000000$, we get $10$ consecutive number as the 'sum of digits' for every $10$ numbers($10000000-10000009$, $10000010-10000019$, ...) and there're two 'multiple of $5$' among any $10$ consecutive numbers.

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