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>.
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== See also ==
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{{AIME box|year=2015|n=I|num-b=2|num-a=4}}
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{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Revision as of 12:24, 20 March 2015

Problem

There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$.

Solution

We call the positive integer mentioned $a$. Then $a^3 = 16p+1$.

$a^3 = 16p+1$

$a^3-1 = 16p$

Factoring the left side:

$(a-1)\cdot(a^2+a+1) = 16p$

We can then try setting one of the factors to $16$, starting with $a-1$.

We get $a=16+1=17$

Then our other factor is $a^2+a+1=17^2+17+1=289+17+1=307$. A quick divisibility search shows that $307$ is prime, so our answer is $\boxed{307}$.

See also

2015 AIME I (ProblemsAnswer KeyResources)
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

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