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Topic: Number with n^3 , 109
Replies: 5   Last Post: Jan 10, 2013 5:42 AM

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AP

Posts: 137
Registered: 3/4/09
Re: Number with n^3 , 109
Posted: Jan 10, 2013 5:42 AM
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On Wed, 2 Jan 2013 00:52:06 +0900, "mina_world"
<mina_world@hanmail.net> wrote:

>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)

a+(a+r)+(a+2r)+(a+3r)+...+b=((b-a)/r+1)(a+b)/2

( (b-a)/r+1 is the number of terms in the sum ; this generalize the
number of terms in
p+(p+1)+(p+2)+...+q : q-p+1)

for example if a=1, r=1 , we obtain b(b+1)/2

so 57+59+....+109=((109-57)/2+1)(57+109)/2=27x83

and 3n^3+9n^2+15n+9=2241
n^3+3n^2+5n=744

so n^3<744 and n<=9
but n divides 744=2^3*3*31
and also, n must be even
so n=2 (false) or 4 (false) and the only possibility is 8 : true



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