Difference between revisions of "2015 AIME I Problems/Problem 3"
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There is a prime number <math>p</math> such that <math>16p+1</math> is the cube of a positive integer. Find <math>p</math>. | There is a prime number <math>p</math> such that <math>16p+1</math> is the cube of a positive integer. Find <math>p</math>. | ||
− | ==Solution== | + | ==Solution 1== |
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. |
Revision as of 12:02, 15 March 2018
Problem
There is a prime number such that is the cube of a positive integer. Find .
Solution 1
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, .
Another Another Solution
Let . Realize that is , so let . Expansion, then division by 4, gets . Clearly for some . Substitution and another division by 4 gets . Since is prime and there is a factor of in the LHS, . 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.