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

Revision as of 01:16, 17 January 2021

Problem

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

Video Solution

~ 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

@@ Line 1: / Line 1: @@
 == Problem ==
 What is that largest [[positive integer]] <math>n</math> for which <math>n^3+100</math> is [[divisible]] by <math>n+10</math>?
+== Video Solution ==
+https://youtu.be/zfChnbMGLVQ?t=1458
+~ pi_is_3.14
 == Solution 1 ==

1986 AIME (Problems • Answer Key • Resources)
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