Junior Divisional 2013/3
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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.
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.
Find out the greatest integer $n$ for which $n^3+500$ will be divisible by $n+10$.
Re: Faridpur, 12th bdmo , j19
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 $490$
$n^{3}+500=n^{3}+10^{3}-500$
$=(n+10)(n^{2}-10n+100)-500$
$(n+10)$ 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 $490$
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