Difference between revisions of "1986 AIME Problems/Problem 5"
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After applying long division, we see that <math>\frac{n^3+100}{n+10} = n^2 - 10n + 100 - \frac{900}{n+10}</math>. Thus, <math>n+10</math> must be a factor of <math>900</math>, and if we want the largest value of <math>n</math>, we have that <math>n+10 = 900 \Longrightarrow n = \boxed{890}</math>. | After applying long division, we see that <math>\frac{n^3+100}{n+10} = n^2 - 10n + 100 - \frac{900}{n+10}</math>. Thus, <math>n+10</math> must be a factor of <math>900</math>, and if we want the largest value of <math>n</math>, we have that <math>n+10 = 900 \Longrightarrow n = \boxed{890}</math>. | ||
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== See also == | == See also == | ||
Revision as of 23:34, 30 January 2018
Contents
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 division. We are looking for 
. Again, 
must be a factor of 
.
Solution 2
After applying long division, we see that 
. Thus, must be a factor of 
, and if we want the largest value of , we have that 
.
-ilovepi3.14
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
