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Difference between revisions of "2015 AIME I Problems/Problem 3"
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==Solution 10 (Lengthy Brute Force)==
 
==Solution 10 (Lengthy Brute Force)==
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Recognizing that AIME answers are <math>0</math> through <math>999</math>, the numbers whose cube could even be in contention to be equal to <math>16p + 1</math> are <math>5-25</math>. The cubes of <math>1-5</math> are all below <math>17</math>. We might consider <math>1</math>, but that would result in a <math>p</math> of <math>0</math>, which is not prime and does not follow our given conditions. Hence, we brute force by looking at all cubes of <math>5-25</math> until we get that the cube of <math>17</math>, <math>4913</math>, works when <math>p=\boxed{307}</math>
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Recognizing that AIME answers are <math>0</math> through <math>999</math>, the numbers whose cube could even be in contention to be equal to <math>16p + 1</math> are <math>4-25</math>. The cubes of <math>1-3</math> are all below <math>17</math>. We might consider <math>1</math>, but that would result in a <math>p</math> of <math>0</math>, which is not prime and does not follow our given conditions. Hence, we brute force by looking at all cubes of <math>5-25</math> until we get that the cube of <math>17</math>, <math>4913</math>, works when <math>p=\boxed{307}</math>
  
 
Solution by: armang32324
 
Solution by: armang32324

Revision as of 19:05, 6 April 2023

Problem

There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$.

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

https://youtu.be/vajttONefxs

~savannahsolver

Solution 1

Let the positive integer mentioned be $a$, so that $a^3 = 16p+1$. Note that $a$ must be odd, because $16p+1$ is odd.

Rearrange this expression and factor the left side (this factoring can be done using $(a^3-b^3) = (a-b)(a^2+a b+b^2)$ or synthetic divison once it is realized that $a = 1$ is a root):

\begin{align*} a^3-1 &= 16p\\ (a-1)(a^2+a+1) &= 16p\\ \end{align*}

Because $a$ is odd, $a-1$ is even and $a^2+a+1$ is odd. If $a^2+a+1$ is odd, $a-1$ must be some multiple of $16$. However, for $a-1$ to be any multiple of $16$ other than $16$ would mean $p$ is not a prime. Therefore, $a-1 = 16$ and $a = 17$.

Then our other factor, $a^2+a+1$, is the prime $p$:

\begin{align*} (a-1)(a^2+a+1) &= 16p\\ (17-1)(17^2+17+1) &=16p\\ p = 289+17+1 &= \boxed{307} \end{align*}

Solution 2 (Similar to 1)

Observe that this is the same as $16p+1=n^3$ for some integer $n$. So:

\begin{align*} 16p &= n^3-1\\ 16p &= n^3-1^3\\ 16p &= (n-1)(n^2+n+1)\\ \end{align*}

Observe that either $p=n-1$ or $p=n^2+n+1$ because $p$ and $16$ share no factors ($p$ can't be $2$). Let $p=n-1$. Then:

\begin{align*} p &= n-1\\ 16 &= n^2+n+1\\ n^2+n &= 15\\ n(n+1) &= 15\\ \end{align*}

Which is impossible for integer n. So $p=n^2+n+1$ and

\begin{align*} 16 &= n-1\\ n &= 17\\ p &= 17^2+17+1\\ p = 289+17+1 &= \boxed{307}\\ \end{align*} - firebolt360

Solution 3

Since $16p+1$ is odd, let $16p+1 = (2a+1)^3$. Therefore, $16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1$. From this, we get $8p=a(4a^2+6a+3)$. We know $p$ is a prime number and it is not an even number. Since $4a^2+6a+3$ is an odd number, we know that $a=8$.

Therefore, $p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}$.

Solution 4

Let $16p+1=a^3$. Realize that $a$ congruent to $1\mod 4$, so let $a=4n+1$. Expansion, then division by 4, gets $16n^3+12n^2+3n=4p$. Clearly $n=4m$ for some $m$. Substitution and another division by 4 gets $256m^3+48m^4+3m=p$. Since $p$ is prime and there is a factor of $m$ in the LHS, $m=1$. Therefore, $p=\boxed{307}$.

Solution 5

Notice that $16p+1$ must be in the form $(a+1)^3 = a^3 + 3a^2 + 3a + 1$. Thus $16p = a^3 + 3a^2 + 3a$, or $16p = a\cdot (a^2 + 3a + 3)$. Since $p$ must be prime, we either have $p = a$ or $a = 16$. Upon further inspection and/or using the quadratic formula, we can deduce $p \neq a$. Thus we have $a = 16$, and $p = 16^2 + 3\cdot 16 + 3 = \boxed{307}$.

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, $p=\boxed{307}$ is our only answer

pi_is_3.141

Solution 7

If $16p+1 =k$, we have $k \equiv 1 \mod 16$, so $k^3 \equiv 1 \mod 16$. If $k=1$ we have $p=0$, which is not prime. If $k=17$ we have $16p+1=4913$, or $p=\boxed{307}$

Solution 8 (Pattern Recognition)

Notice that: \begin{align*} 1^3 &= 0 +1\\ 2^3 &= 1*7+1 \\ 3^3 &= 2*13+1\\ 4^3 &= 3*21+1\\ 5^3 &= 4*31+1\\ 6^3 &= 5*43+1 \end{align*}

Here, we can see a clear pattern that $n^3=(n-1)p+1$, where $p$ is some positive (not necessarily prime) integer. Hence, the equation $16p+1=a^3$ can interpret as $17^3 = 16p+1$. Solving it, we got $p=307$. After checking all possible divisors, we will find that $307$ is prime. Hence, we got $p=\boxed{307}$.

Solution 9 (Slightly Different Modular Arithmetic)

We see that $16p+1=n^3$ for a positive integer $n$. Subtracting $1$, we can turn this equation into a modular congruence, since $n^3-1$ must be a multiple of $16$.

Since $n^3-1\equiv0\pmod{16}$, $n^3\equiv1\pmod{16}$. We observe that $n=1$ is a solution to this congruence, which doesn't work. The next, or most obvious number to try is $n=17$. Plugging this in to our original equation, we get

$17^3=16p+1$, yielding $p=\boxed{307}$, which is prime.


-among us (countmath1)

Solution 10 (Lengthy Brute Force)

Recognizing that AIME answers are $0$ through $999$, the numbers whose cube could even be in contention to be equal to $16p + 1$ are $4-25$. The cubes of $1-3$ are all below $17$. We might consider $1$, but that would result in a $p$ of $0$, which is not prime and does not follow our given conditions. Hence, we brute force by looking at all cubes of $5-25$ until we get that the cube of $17$, $4913$, works when $p=\boxed{307}$

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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