Difference between revisions of "1986 AIME Problems/Problem 5"

(Solution 2)
(Solution 2)
Line 6: Line 6:
 
In a similar manner, we can apply synthetic division. We are looking for <math>\frac{n^3 + 100}{n + 10} = n^2 - 10n - 100 - \frac{900}{n + 10}</math>. Again, <math>n + 10</math> must be a factor of <math>900 \Longrightarrow n = \boxed{890}</math>.
 
In a similar manner, we can apply synthetic division. We are looking for <math>\frac{n^3 + 100}{n + 10} = n^2 - 10n - 100 - \frac{900}{n + 10}</math>. Again, <math>n + 10</math> must be a factor of <math>900 \Longrightarrow n = \boxed{890}</math>.
  
 
== Solution 2 ==
 
 
After applying long division, we see that <math>\frac{n^3+100}{n+10} = n^2 - 10n + 100 - \frac{900}{n+10}</math>. Thus, <math>n+10</math> must be a factor of <math>900</math>, and if we want the largest value of <math>n</math>, we have that <math>n+10 = 900 \Longrightarrow n = \boxed{890}</math>.
 
 
-ilovepi3.14
 
  
 
== See also ==
 
== See also ==

Revision as of 00:34, 31 January 2018

Problem

What is that largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?

Solution

If $n+10 \mid n^3+100$, $\gcd(n^3+100,n+10)=n+10$. Using the Euclidean algorithm, we have $\gcd(n^3+100,n+10)= \gcd(-10n^2+100,n+10)$ $= \gcd(100n+100,n+10)$ $= \gcd(-900,n+10)$, so $n+10$ must divide 900. The greatest integer $n$ for which $n+10$ divides 900 is 890; we can double-check manually and we find that indeed $900 \mid 890^3+100$.

In a similar manner, we can apply synthetic division. We are looking for $\frac{n^3 + 100}{n + 10} = n^2 - 10n - 100 - \frac{900}{n + 10}$. Again, $n + 10$ must be a factor of $900 \Longrightarrow n = \boxed{890}$.


See also

1986 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS