Another (easy) problem for young people
I'm going to propose you another pretty exercise which can be done by everybody (Steven, this is for you! 
).
Find the largest positive integer $n$ for which
$n+10|n^3+100$.
I have solved it, but I am not very sure of the solution I've found. If you wanted to try, I would be grateful.
). Find the largest positive integer $n$ for which
$n+10|n^3+100$.
I have solved it, but I am not very sure of the solution I've found. If you wanted to try, I would be grateful.
Risposte
"Steven":
Of course
Thanks for your correction, I edited the message.
See you soon!
You are welcome. Night.
Thanks.
Of course
Thanks for your correction, I edited the message.
See you soon!
Thanks for your correction, I edited the message.
See you soon!
Great, my dear friend, as always.
Just a very very little mistake: pay attention to the divisor (I've written $n+10$ while you wrote $n+1$... you simply forgot some $0$...)
I got $890$.
Thanks for answering.
Enjoy your evening too.
Just a very very little mistake: pay attention to the divisor (I've written $n+10$ while you wrote $n+1$... you simply forgot some $0$...
)I got $890$.
Thanks for answering.
Enjoy your evening too.
Here I am 
We are looking for the largest $n$ which lets that expression be integer.
It results
$\frac{n^3+100}{n+10}=\frac{n+10^3-900}{n+10}=frac{(n+10)(n^2+100-10n)-900}{n+10}=n^2+100-10n-\frac{900}{n+10}$
So we have to impose that $n+10|900$ and we obtain the major $n$ when
$n+10=900\quad\quad=>\quad\quad n=890$
Did you get the same number?
Bye, have a good evening.
We are looking for the largest $n$ which lets that expression be integer.
It results
$\frac{n^3+100}{n+10}=\frac{n+10^3-900}{n+10}=frac{(n+10)(n^2+100-10n)-900}{n+10}=n^2+100-10n-\frac{900}{n+10}$
So we have to impose that $n+10|900$ and we obtain the major $n$ when
$n+10=900\quad\quad=>\quad\quad n=890$
Did you get the same number?
Bye, have a good evening.
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