Last goes first

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Raiyan Jamil
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Last goes first

Unread post by Raiyan Jamil » Sun Feb 09, 2014 9:01 pm

In a six digit number , the rightmost digit is 2 . If this 2 is removed from its place and is placed in the leftmost place , then a new six digit number is made . If this number is one third of the number before , what is the new number ?
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Fatin Farhan
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Re: Last goes first

Unread post by Fatin Farhan » Mon Feb 10, 2014 11:07 am

$$10^5a+10^4b+10^3c+10^2d+10e+2=$$ $$10^5*6+10^4*3a+10^3*3b+10^2*3c+10*3d+3e$$
Solving this you will get the previous number $$857142$$ and the new number $$\boxed{285714}$$.
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SANZEED
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Re: Last goes first

Unread post by SANZEED » Mon Feb 10, 2014 6:10 pm

@Fatin:
It is easier to solve the equation, $10a+2=3\times (200000+a)$.
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Tahmid
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Re: Last goes first

Unread post by Tahmid » Mon Feb 10, 2014 7:45 pm

solution is same as @sanzeed vai
let our number is $10a+2$ here a is a five digit number
if 2 is placed at the leftmost then the number becomes $(2*10^{5})+a\Rightarrow 200000+a$

so, $10a+2=3(200000+a)$
after solving this equation we get $a=85714$
ans is $200000+a=200000+85714=285714$ :)

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