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

(See also)
m (breakup)
Line 2: Line 2:
 
What is that largest [[positive integer]] <math>n</math> for which <math>n^3+100</math> is [[divisible]] by <math>n+10</math>?
 
What is that largest [[positive integer]] <math>n</math> for which <math>n^3+100</math> is [[divisible]] by <math>n+10</math>?
 
== Solution ==
 
== Solution ==
If <math>n+10 \mid n^3+100</math>, <math>\gcd(n^3+100,n+10)=n+10</math>. Using the [[Euclidean algorithm]], we have <math>\gcd(n^3+100,n+10)= \gcd(-10n^2+100,n+10)= \gcd(100n+100,n+10)= \gcd(-900,n+10)</math>, so <math>n+10</math> must divide 900. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides 900 is 890; we can double-check manually and we find that indeed <math>900 \mid 890^3+100</math>.
+
If <math>n+10 \mid n^3+100</math>, <math>\gcd(n^3+100,n+10)=n+10</math>. Using the [[Euclidean algorithm]], we have <math>\gcd(n^3+100,n+10)= \gcd(-10n^2+100,n+10)</math> <math>= \gcd(100n+100,n+10)</math> <math>= \gcd(-900,n+10)</math>, so <math>n+10</math> must divide 900. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides 900 is 890; we can double-check manually and we find that indeed <math>900 \mid 890^3+100</math>.
  
In a similar manner, we can apply [[synthetic substitution]]. 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 = 890</math>.
+
In a similar manner, we can apply [[synthetic substitution]]. 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>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1986|num-b=4|num-a=6}}
 
{{AIME box|year=1986|num-b=4|num-a=6}}
* [[AIME Problems and Solutions]]
 
* [[American Invitational Mathematics Examination]]
 
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 20:21, 9 April 2008

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 substitution. 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
Invalid username
Login to AoPS