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Last goes first
Posted: Sun Feb 09, 2014 9:01 pm
by Raiyan Jamil
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 ?
Re: Last goes first
Posted: Mon Feb 10, 2014 11:07 am
by Fatin Farhan
$$10^5a+10^4b+10^3c+10^2d+10e+2=$$ $$10^5*6+10^4*3a+10^3*3b+10^2*3c+10*3d+3e$$
Solving this you will get the previous number $$857142$$ and the new number $$\boxed{285714}$$.
Re: Last goes first
Posted: Mon Feb 10, 2014 6:10 pm
by SANZEED
@Fatin:
It is easier to solve the equation, $10a+2=3\times (200000+a)$.
Re: Last goes first
Posted: Mon Feb 10, 2014 7:45 pm
by Tahmid
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 $200000+a=200000+85714=285714$
