Bdmo 2014 national: junior 04

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atiab jobayer
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Bdmo 2014 national: junior 04

Unread post by atiab jobayer » Fri Feb 21, 2014 6:29 pm

The unit digit of a six digit number is 1 and it is removed, leaving a five digit number. The removed unit digit is 1 is then placed at the far left of the five digit number,making a new six digit number. If the new number is 3/1 of the original number,what is the original number?
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Tahmid
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Re: Bdmo 2014 national: junior 04

Unread post by Tahmid » Sat Feb 22, 2014 12:46 am

let, $x$ is a 5 digit number. If we place 1 at the right most of $x$ we get the original 6 digit number.
so the original 6 digit number is $10x+1$
again if we place 1 at the left most of $x$ we get the new 6 digit number.
so the new 6 digit number is ${10}^{5}+x$

given the new number is $\frac{1}{3}$ of the original number.

so,${10}^{5}+x=\frac{1}{3}(10x+1)$
$\Leftrightarrow (3*{10}^{5})+3x=10x+1$
$\Leftrightarrow 7x=300000-1$
$\Leftrightarrow 7x=299999$
$\Leftrightarrow x=48257$

so, the original number is $10x+1=(10*48257)+1=482571$ [ans] :)

Kiriti
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Re: Bdmo 2014 national: junior 04

Unread post by Kiriti » Sat Feb 22, 2014 11:52 am

এই প্রবলেমটা অনেকটা IMO-1962-1 এর মত । :D
"Education is the most powerful weapon which you can use to change the world"- Nelson Mandela

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atiab jobayer
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Re: Bdmo 2014 national: junior 04

Unread post by atiab jobayer » Sun Feb 23, 2014 10:52 am

Tahmid wrote:let, $x$ is a 5 digit number. If we place 1 at the right most of $x$ we get the original 6 digit number.
so the original 6 digit number is $10x+1$
again if we place 1 at the left most of $x$ we get the new 6 digit number.
so the new 6 digit number is ${10}^{5}+x$

given the new number is $\frac{1}{3}$ of the original number.

so,${10}^{5}+x=\frac{1}{3}(10x+1)$
$\Leftrightarrow (3*{10}^{5})+3x=10x+1$
$\Leftrightarrow 7x=300000-1$
$\Leftrightarrow 7x=299999$
$\Leftrightarrow x=48257$

so, the original number is $10x+1=(10*48257)+1=482571$ [ans] :)
Can you give the answer in bengali?
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Tahmid
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Re: Bdmo 2014 national: junior 04

Unread post by Tahmid » Sun Feb 23, 2014 12:37 pm

অবশ্যই পারি,
ধরি, $x$ একটি 5 অংকের সংখ্যা যার শেষে [সর্বডানে] 1 বসালে আমরা আসল 6 অংকের সংখ্যাটি পাই
সুতরাং, আসল 6 অংকের সংখ্যাটি হল is $10x+1$
আবার $x$ এর সামনে [সর্ববামে] 1 বসালে আমরা নতুন 6 অংকের সংখ্যাটি পাই
সুতরাং, নতুন 6 অংকের সংখ্যাটি হল ${10}^{5}+x$

দেয়া আছে নতুন সংখ্যাটি আসল সংখ্যার $\frac{1}{3}$ অংশ

অতএব,${10}^{5}+x=\frac{1}{3}(10x+1)$
$\Leftrightarrow (3*{10}^{5})+3x=10x+1$
$\Leftrightarrow 7x=300000-1$
$\Leftrightarrow 7x=299999$
$\Leftrightarrow x=48257$

সুতরাং আসল সংখ্যাটি হল $10x+1=(10*48257)+1=482571$ [ans] :)

[Moderator edit: the rule requires you to use English in here http://www.matholympiad.org.bd/forum/viewforum.php?f=24 , not Olympiad and Other Programs froum])

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