2015 AIME I Problems/Problem 3
Contents
- 1 Problem
- 2 Video Solution For Problems 1-3
- 3 Video Solution by OmegaLearn
- 4 Video Solution by WhyMath
- 5 Solution 1
- 6 Solution 2 (Similar to 1)
- 7 Solution 3
- 8 Solution 4
- 9 Solution 5
- 10 Solution 6
- 11 Solution 7
- 12 Solution 8 (Pattern Recognition)
- 13 Solution 9 (Slightly Different Modular Arithmetic)
- 14 Solution 10 (Lengthy Brute Force)
- 15 See also
Problem
There is a prime number such that is the cube of a positive integer. Find .
Video Solution For Problems 1-3
https://www.youtube.com/watch?v=5HAk-6qlOH0
Video Solution by OmegaLearn
https://youtu.be/3bRjcrkd5mQ?t=1096
~ pi_is_3.14
Video Solution by WhyMath
~savannahsolver
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 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
Solution 7
If , we have , so . If we have , which is not prime. If we have , or
Solution 8 (Pattern Recognition)
Notice that:
Here, we can see a clear pattern that , where is some positive (not necessarily prime) integer. Hence, the equation can interpret as . Solving it, we got . After checking all possible divisors, we will find that is prime. Hence, we got .
Solution 9 (Slightly Different Modular Arithmetic)
We see that for a positive integer . Subtracting , we can turn this equation into a modular congruence, since must be a multiple of .
Since , . We observe that is a solution to this congruence, which doesn't work. The next, or most obvious number to try is . Plugging this in to our original equation, we get
, yielding , which is prime.
-among us (countmath1)
Solution 10 (Lengthy Brute Force)
Recognizing that AIME answers are through , the numbers whose cube could even be in contention to be equal to are . The cubes of are all below . We might consider , but that would result in a of , which is not prime and does not follow our given conditions. Hence, we brute force by looking at all cubes of until we get that the cube of , , works when
Solution by: armang32324
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 |
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