Junior Divisional 2013/3

Problem for Junior Group from Divisional Mathematical Olympiad will be solved here.
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Please don't post problems (by starting a topic) in the "Junior: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
barnik
Posts: 13
Joined: Wed Dec 03, 2014 3:37 pm

Junior Divisional 2013/3

Unread post by barnik » Tue Dec 09, 2014 8:05 pm

Find out the greatest integer $n$ for which $n^3+500$ will be divisible by $n+10$.

tanmoy
Posts: 282
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

Re: Faridpur, 12th bdmo , j19

Unread post by tanmoy » Wed Dec 10, 2014 2:39 pm

Please,use $Latex$ in writing equations so that the equations look beautiful.BTW,here is the solution:
$n^{3}+500=n^{3}+10^{3}-500$
$=(n+10)(n^{2}-10n+100)-500$
$(n+10)$ divides $(n+10)(n^{2}-10n+100)$.So,$(n+10)$ must divide $500$.The greatest value of $(n+10)$ which divides $500$ is $500$.$\therefore$ the greatest value of $n$ is $490$ :)
"Questions we can't answer are far better than answers we can't question"

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