Difference between revisions of "1986 AIME Problems/Problem 5"
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~qwertysri987 | ~qwertysri987 | ||
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+ | ==Solution 5 (Easy Modular Arithmetic)== | ||
+ | Notice that <math>n\equiv -10 \pmod{n+10}</math>. Therefore | ||
+ | <cmath> | ||
+ | 0 \equiv n^3+100\equiv(-10)^3+100=-900 \pmod{n+10} \Rightarrow n+10 | 900 \Rightarrow \max_{n+10 | n^3+100} {n} = \boxed{890}. | ||
+ | </cmath> | ||
+ | |||
+ | ~asops | ||
== See also == | == See also == |
Revision as of 10:48, 15 September 2021
Contents
Problem
What is that largest positive integer for which
is divisible by
?
Video Solution
https://youtu.be/zfChnbMGLVQ?t=1458
~ pi_is_3.14
~ momeme
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 (Simple)
Let , then
. Then
Therefore,
must be divisible by
, which is largest when
and
Solution 3
In a similar manner, we can apply synthetic division. We are looking for . Again,
must be a factor of
.
Solution 4
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
Solution 5 (Easy Modular Arithmetic)
Notice that . Therefore
~asops
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.