Difference between revisions of "1986 AIME Problems/Problem 5"
I_like_pie (talk | contribs) (→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}} | ||
− | |||
− | |||
− | |||
[[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 for which is divisible by ?
Solution
If , . Using the Euclidean algorithm, we have , so must divide 900. The greatest integer for which divides 900 is 890; we can double-check manually and we find that indeed .
In a similar manner, we can apply synthetic substitution. We are looking for . Again, must be a factor of .
See also
1986 AIME (Problems • Answer Key • Resources) | ||
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 |