Difference between revisions of "2015 AIME I Problems/Problem 3"
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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. | 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. | ||
− | 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> | + | 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): |
− | < | + | <cmath>\begin{align*} |
− | + | a^3-1 &= 16p\\ | |
− | + | (a-1)(a^2+a+1) &= 16p\\ | |
+ | \end{align*}</cmath> | ||
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>. | 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>. | ||
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Then our other factor, <math>a^2+a+1</math>, is the prime <math>p</math>: | Then our other factor, <math>a^2+a+1</math>, is the prime <math>p</math>: | ||
− | < | + | <cmath>\begin{align*} |
− | + | (a-1)(a^2+a+1) &= 16p\\ | |
− | + | (17-1)(17^2+17+1) &=16p\\ | |
− | + | p = 289+17+1 &= \boxed{307} | |
− | + | \end{align*}</cmath> | |
==Another Solution== | ==Another Solution== | ||
− | Since <math>16p+1</math> is odd, let <math>16p+1 = (2a+1)^3</math> | + | Since <math>16p+1</math> is odd, let <math>16p+1 = (2a+1)^3</math>. Therefore, <math>16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1</math>. From this, we get <math>8p=a(4a^2+6a+3)</math>. We know <math>p</math> is a prime number and it is not an even number. Since <math>4a^2+6a+3</math> is an odd number, we know that <math>a=8</math>. |
− | |||
− | <math>16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1</math> | ||
− | |||
− | |||
− | |||
− | <math>8p=a(4a^2+6a+3)</math> | ||
− | |||
− | We know p is a prime number and | ||
− | |||
− | + | Therefore, <math>p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}</math>. | |
== See also == | == See also == |
Revision as of 13:27, 21 February 2017
Contents
Problem
There is a prime number such that is the cube of a positive integer. Find .
Solution
Let the positive integer mentioned be , so that . Note that must be odd, because is odd.
Rearrange this expression and factor the left side (this factoring can be done using or synthetic divison once it is realized that is a root):
Because is odd, is even and is odd. If is odd, must be some multiple of . However, for to be any multiple of other than would mean is not a prime. Therefore, and .
Then our other factor, , is the prime :
Another Solution
Since is odd, let . Therefore, . From this, we get . We know is a prime number and it is not an even number. Since is an odd number, we know that .
Therefore, .
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.