divide it

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Rafe
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divide it

Unread post by Rafe » Thu Dec 27, 2012 6:38 pm

show that the number abbcca is divided by 11 for any value of a,b,c within 1-9
Last edited by Rafe on Sat Dec 29, 2012 11:30 am, edited 1 time in total.

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SANZEED
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Re: divide it

Unread post by SANZEED » Thu Dec 27, 2012 8:00 pm

Rafe wrote:show that the number abbcca is divided by 7 for any value of a,b,c within 1-9
I think this statement is incorrect. For a counter example is $211332$ where $a=2,b=1,c=3$.
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Fahim Shahriar
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Re: divide it

Unread post by Fahim Shahriar » Thu Dec 27, 2012 8:52 pm

The statement is not true at all. If you said $abcabc$, then it would be divisible by $7$.
Name: Fahim Shahriar Shakkhor
Notre Dame College

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SANZEED
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Re: divide it

Unread post by SANZEED » Fri Dec 28, 2012 11:08 am

Fahim Shahriar wrote:The statement is not true at all. If you said $abcabc$, then it would be divisible by $7$.
Again,with the criteria of divisibility by $7$, we can also show that the first number $abbcca$ is dividible by $7$, if the number $3c-2a-b$ is divisible by $7$. That's how I find examples like $7|211442$ and counter-example $211332$.
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Rafe
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Joined: Wed Oct 24, 2012 11:25 am

Re: divide it

Unread post by Rafe » Sat Dec 29, 2012 11:33 am

sorry friends.i have corrected that

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Phlembac Adib Hasan
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Re: divide it

Unread post by Phlembac Adib Hasan » Sat Dec 29, 2012 11:58 am

Rafe wrote:show that the number abbcca is divided by 11 for any value of a,b,c within 1-9
$a-b+b-c+c-a=0\equiv 0(\bmod \; 11)$ so by the divisibility rule of $11$ we can conclude that $(abbcca)_{10}\equiv 0(\bmod \; 11)$
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