Difference between revisions of "2015 AIME I Problems/Problem 3"

(Solution)
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==Solution==
 
==Solution==
  
We call the positive integer mentioned <math>a</math>.  Then <math>a^3 = 16p+1</math>.
+
Let the positive integer mentioned be <math>a</math>, so that <math>a^3 = 16p+1</math>. Note that <math>a</math> must be odd, because <math>16p+1</math> is odd.
  
<math>a^3 = 16p+1</math>
+
Rearrange this expression and factor the left side (this factoring can be done using <math>(a^3-b^3) = (a-b)(a^2+a b+b^2)</math>, or synthetic divison once it is realized that <math>a = 1</math> is a root):
  
 
<math>a^3-1 = 16p</math>
 
<math>a^3-1 = 16p</math>
  
Factoring the left side:
+
<math>(a-1)(a^2+a+1) = 16p</math>
  
<math>(a-1)\cdot(a^2+a+1) = 16p</math>
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Because <math>a</math> is odd, <math>a-1</math> is even and <math>a^2+a+1</math> is odd. If <math>a^2+a+1</math> is odd, <math>a-1</math> must be some multiple of <math>16</math>. However, for <math>a-1</math> to be any multiple of <math>16</math> other than <math>16</math> would mean <math>p</math> is not a prime. Therefore, <math>a-1 = 16</math> and <math>a = 17</math>.
  
We can then try setting one of the factors to <math>16</math>, starting with <math>a-1</math>.
+
Then our other factor, <math>a^2+a+1</math>, is the prime <math>p</math>:
  
We get <math>a=16+1=17</math>
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<math>(a-1)(a^2+a+1) = 16p</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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<math>(17-1)(17^2+17+1) =16p</math>
 +
 
 +
<math>p = 289+17+1 = \boxed{307}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:14, 20 March 2015

Problem

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

Solution

Let the positive integer mentioned be $a$, so that $a^3 = 16p+1$. Note that $a$ must be odd, because $16p+1$ is odd.

Rearrange this expression and factor the left side (this factoring can be done using $(a^3-b^3) = (a-b)(a^2+a b+b^2)$, or synthetic divison once it is realized that $a = 1$ is a root):

$a^3-1 = 16p$

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

Because $a$ is odd, $a-1$ is even and $a^2+a+1$ is odd. If $a^2+a+1$ is odd, $a-1$ must be some multiple of $16$. However, for $a-1$ to be any multiple of $16$ other than $16$ would mean $p$ is not a prime. Therefore, $a-1 = 16$ and $a = 17$.

Then our other factor, $a^2+a+1$, is the prime $p$:

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

$(17-1)(17^2+17+1) =16p$

$p = 289+17+1 = \boxed{307}$.

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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