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

Revision as of 13:40, 6 May 2007

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 $900 \Longrightarrow n = 890$ .

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 == See also ==
-* [[1986 AIME Problems]]
 {{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]]

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

Revision as of 13:40, 6 May 2007

Problem

Solution

See also