Difference between revisions of "1986 AIME Problems/Problem 5"
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== Solution 2 == | == Solution 2 == | ||
− | In a similar manner, we can apply synthetic division. We are looking for <math>\frac{n^3 + 100}{n + 10} = n^2 - 10n | + | 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 3== | ==Solution 3== |
Revision as of 23:17, 16 February 2020
Problem
What is that largest positive integer for which
is divisible by
?
Solution 1
If ,
. Using the Euclidean algorithm, we have
, so
must divide
. The greatest integer
for which
divides
is
; we can double-check manually and we find that indeed
.
Solution 2
In a similar manner, we can apply synthetic division. We are looking for . Again,
must be a factor of
.
Solution 3
The key to this problem is to realize that for all
. Since we are asked to find the maximum possible
such that
, we have:
. This is because of the property that states that if
and
, then
. Since, the largest factor of 900 is itself we have:
~qwertysri987
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 |
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