Difference between revisions of "2015 AIME I Problems/Problem 3"
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Then our other factor is <math>a^2+a+1=17^2+17+1=289+17+1=307</math>. A quick divisibility search shows that <math>307</math> is prime, so our answer is <math>\boxed{307}</math>. | Then our other factor is <math>a^2+a+1=17^2+17+1=289+17+1=307</math>. A quick divisibility search shows that <math>307</math> is prime, so our answer is <math>\boxed{307}</math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2015|n=I|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Revision as of 13:24, 20 March 2015
Problem
There is a prime number such that
is the cube of a positive integer. Find
.
Solution
We call the positive integer mentioned . Then
.
Factoring the left side:
We can then try setting one of the factors to , starting with
.
We get
Then our other factor is . A quick divisibility search shows that
is prime, so our answer is
.
See also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.