Difference between revisions of "2015 AIME I Problems/Problem 3"
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Notice that <math>16p+1</math> must be in the form <math>(a+1)^3 = a^3 + 3a^2 + 3a + 1</math>. Thus <math>16p = a^3 + 3a^2 + 3a</math>, or <math>16p = a\cdot (a^2 + 3a + 3)</math>. Since <math>p</math> must be prime, we either have <math>p = a</math> or <math>a = 16</math>. Upon further inspection and/or using the quadratic formula, we can deduce <math>p \neq a</math>. Thus we have <math>a = 16</math>, and <math>p = 16^2 + 3\cdot 16 + 3 = \boxed{307}</math>. | Notice that <math>16p+1</math> must be in the form <math>(a+1)^3 = a^3 + 3a^2 + 3a + 1</math>. Thus <math>16p = a^3 + 3a^2 + 3a</math>, or <math>16p = a\cdot (a^2 + 3a + 3)</math>. Since <math>p</math> must be prime, we either have <math>p = a</math> or <math>a = 16</math>. Upon further inspection and/or using the quadratic formula, we can deduce <math>p \neq a</math>. Thus we have <math>a = 16</math>, and <math>p = 16^2 + 3\cdot 16 + 3 = \boxed{307}</math>. | ||
+ | ==Solution 5== | ||
+ | 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, <math>p=\boxed{307}</math> is our only answer | ||
== See also == | == See also == | ||
{{AIME box|year=2015|n=I|num-b=2|num-a=4}} | {{AIME box|year=2015|n=I|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] |
Revision as of 01:23, 8 March 2020
Contents
[hide]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
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 3
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 4
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 5
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
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.