Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Another Solution)
m (→‎Another Solution)
Line 29: Line 29:
 
<math>16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1</math>
 
<math>16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1</math>
  
−
We got:
+
We get:
  
 
<math>8p=a(4a^2+6a+3)</math>
 
<math>8p=a(4a^2+6a+3)</math>

Revision as of 08:19, 30 April 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 $p$:

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

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

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

Another Solution

Since $16p+1$ is odd, let $16p+1 = (2a+1)^3$

$16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1$

We get:

$8p=a(4a^2+6a+3)$

We know p is a prime number and apparently not an even number. and $4a^2+6a+3$ is an odd number, so a must equal 8.

so we get $p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_3&oldid=70071"