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Re: Number with n^3 , 109
Posted:
Jan 1, 2013 11:39 AM
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"mina_world" <mina_world@hanmail.net> wrote in news:kbv0n8$unt$1
@news2.kornet.net:
> Hello, teacher~
>
> n^3 + (n+1)^3 + (n+2)^3 = 57 + 59 + 61 + ... + 109
>
> Find the "n".
>
> ---------------------------------------------------------------
---------
> Answer is 8.
>
> http://board-2.blueweb.co.kr/user/math565/data/math/n3.jpg
>
> This is a solution of elementary schoolchild(of course, clever)
>
> Can you understand it ? I need your explanation.(used formula
etc)
>
I would substitue k=n-1 in the left side. Then when you
expand and cancel, the left side is 3k^3+6k. The sum of the
first M odd numbers is M^2. So the right side is 55^2 - 28^2
= (55-28)(55+28)=27*83.
Divide both sides by 3 and factor the left to get:
k(k^2+2) = 83*9.
So k^2+2 = 83 and k=9 is the obvious solution. So n=k-1=9.
--
Cheerfully resisting change since 1959.
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