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

(See also)
(Solution 1)
 
(20 intermediate revisions by 14 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
What is that largest [[positive integer]] <math>n</math> for which <math>n^3+100</math> is [[divisible]] by <math>n+10</math>?
  
== Solution ==
+
== Video Solution ==
 +
https://youtu.be/zfChnbMGLVQ?t=1458
 +
 
 +
~ pi_is_3.14
 +
 
 +
== Solution 1 ==
 +
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 <math>900</math>. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides <math>900</math> is <math>\boxed{890}</math>; we can double-check manually and we find that indeed <math>900\mid 890^3+100</math>.
 +
 
 +
== Solution 2 (Simple) ==
 +
Let <math>n+10=k</math>, then <math>n=k-10</math>. Then <math>n^3+100 = k^3-30k^2+300k-900</math> Therefore, <math>900</math> must be divisible by <math>k</math>, which is largest when <math>k=900</math> and <math>n=\boxed{890}</math>
 +
 
 +
== Solution 3 ==
 +
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 4==
 +
The key to this problem is to realize that <math>n+10 \mid n^3 +1000</math> for all <math>n</math>. Since we are asked to find the maximum possible <math>n</math> such that <math>n+10 \mid n^3 +100</math>, we have: <math>n+10 \mid ((n^3 +1000) - (n^3 +100) \longrightarrow n+10 \mid 900</math>. This is because of the property that states that if <math>a \mid b</math> and <math>a \mid c</math>, then <math>a \mid b \pm c</math>. Since, the largest factor of 900 is itself we have: <math>n+10=900 \Longrightarrow \boxed{n = 890}</math>
 +
 
 +
~qwertysri987
  
 
== See also ==
 
== See also ==
* [[1986 AIME Problems]]
+
{{AIME box|year=1986|num-b=4|num-a=6}}
 +
 
 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 02:16, 17 January 2021

Problem

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

Video Solution

https://youtu.be/zfChnbMGLVQ?t=1458

~ pi_is_3.14

Solution 1

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 $\boxed{890}$; we can double-check manually and we find that indeed $900\mid 890^3+100$.

Solution 2 (Simple)

Let $n+10=k$, then $n=k-10$. Then $n^3+100 = k^3-30k^2+300k-900$ Therefore, $900$ must be divisible by $k$, which is largest when $k=900$ and $n=\boxed{890}$

Solution 3

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 $900 \Longrightarrow n = \boxed{890}$.

Solution 4

The key to this problem is to realize that $n+10 \mid n^3 +1000$ for all $n$. Since we are asked to find the maximum possible $n$ such that $n+10 \mid n^3 +100$, we have: $n+10 \mid ((n^3 +1000) - (n^3 +100) \longrightarrow n+10 \mid 900$. This is because of the property that states that if $a \mid b$ and $a \mid c$, then $a \mid b \pm c$. Since, the largest factor of 900 is itself we have: $n+10=900 \Longrightarrow \boxed{n = 890}$

~qwertysri987

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