Difference between revisions of "2015 AIME I Problems/Problem 3"
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Observe that this is the same as <math>16p+1=n^3</math> for some integer <math>n</math>. | Observe that this is the same as <math>16p+1=n^3</math> for some integer <math>n</math>. | ||
So: | So: | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
16p &= n^3-1\\ | 16p &= n^3-1\\ | ||
Line 33: | Line 34: | ||
16p &= (n-1)(n^2+n+1)\\ | 16p &= (n-1)(n^2+n+1)\\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
Observe that either <math>p=n-1</math> or <math>p=n^2+n+1</math> because <math>p</math> and <math>16</math> share no factors (<math>p</math> can't be <math>2</math>). | Observe that either <math>p=n-1</math> or <math>p=n^2+n+1</math> because <math>p</math> and <math>16</math> share no factors (<math>p</math> can't be <math>2</math>). | ||
Let <math>p=n-1</math>. | Let <math>p=n-1</math>. | ||
Then: | Then: | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
p &= n-1\\ | p &= n-1\\ | ||
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n(n+1) &= 15\\ | n(n+1) &= 15\\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
Which is obviously impossible for integer n. So <math>p=n^2+n+1</math> and | Which is obviously impossible for integer n. So <math>p=n^2+n+1</math> and | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
16 &= n-1\\ | 16 &= n-1\\ |
Revision as of 12:50, 30 May 2020
Contents
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 :
Solution 2 (Similar to 1)
Observe that this is the same as for some integer . So:
Observe that either or because and share no factors ( can't be ). Let . Then:
Which is obviously impossible for integer n. So and
- firebolt360
Solution 3
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, .
Solution 4
Let . Realize that congruent to , 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, .
Solution 5
Notice that must be in the form . Thus , or . Since must be prime, we either have or . Upon further inspection and/or using the quadratic formula, we can deduce . Thus we have , and .
Solution 6
Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. Case one: The cube is of the form 16k+1-->Plugging this in, and taking note that p is prime and has only 1 factor gives p=307 Case two: The cube is of the form 16k+15--> Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here
Hence, is our only answer
pi_is_3.141
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.