2015 AIME I Problems/Problem 3: Difference between revisions

Revision as of 12:24, 20 March 2015

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

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 $\boxed{307}$ .

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

2015 AIME I (Problems • Answer Key • Resources)
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