Difference between revisions of "2019 AIME I Problems/Problem 14"

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==Solution 1==
 
==Solution 1==
  
The problem tells us that <math>2019^8 \equiv -1 \pmod{p}</math> for some prime <math>p</math>. We want to find the smallest odd possible value of <math>p</math>. By squaring both sides of the congruence, we get <math>2019^{16} \equiv 1 \pmod{p}</math>.  
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The problem tells us that <math>2019^8 \equiv -1 \pmod{p}</math> for some prime <math>p</math>. We want to find the smallest odd possible value of <math>p</math>. By squaring both sides of the congruence, we get <math>2019^{16} \equiv 1 \pmod{p}</math>.  
  
By Euler's theorem, <math>2019^{\phi(p)} \equiv 1 \pmod{p}</math>. We also know that <math>ord_p(2019)  \vert  \phi(p)</math>.
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Since <math>2019^{16} \equiv 1 \pmod{p}</math>, <math>ord_p(2019)</math> = <math>1, 2, 4, 8,</math> or <math>16</math>
 
 
Therefore, <math>ord_p(2019)</math> = <math>1, 2, 4, 8,</math> or <math>16</math>
 
  
 
However, if <math>ord_p(2019)</math> = <math>1, 2, 4,</math> or <math>8,</math> then <math>2019^8</math> clearly will be <math>1 \pmod{p} </math> instead of <math>-1 \pmod{p}</math>, causing a contradiction.
 
However, if <math>ord_p(2019)</math> = <math>1, 2, 4,</math> or <math>8,</math> then <math>2019^8</math> clearly will be <math>1 \pmod{p} </math> instead of <math>-1 \pmod{p}</math>, causing a contradiction.
  
Therefore, <math>ord_p(2019) = 16</math>, and <math>\phi(p)</math> is a multiple of 16. Since we know <math>p</math> is prime, <math>\phi(p) = p(1 - \frac{1}{p})</math> or <math>p - 1</math>. Therefore, <math>p</math> must be <math>1 \pmod{16}</math>. The two smallest primes that are <math>1 \pmod{16}</math> are <math>17</math> and <math>97</math>. <math>2019^8 \not\equiv -1 \pmod{17}</math>, but <math>2019^8 \equiv -1 \pmod{97}</math>, so our answer is <math>\boxed{097}</math>.
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Therefore, <math>ord_p(2019) = 16</math>. Because <math>ord_p(2019)  \vert  \phi(p)</math>, <math>\phi(p)</math> is a multiple of 16. Since we know <math>p</math> is prime, <math>\phi(p) = p(1 - \frac{1}{p})</math> or <math>p - 1</math>. Therefore, <math>p</math> must be <math>1 \pmod{16}</math>. The two smallest primes that are <math>1 \pmod{16}</math> are <math>17</math> and <math>97</math>. <math>2019^8 \not\equiv -1 \pmod{17}</math>, but <math>2019^8 \equiv -1 \pmod{97}</math>, so our answer is <math>\boxed{097}</math>.
  
 
===Note to solution 1===
 
===Note to solution 1===

Revision as of 22:19, 16 March 2019

Problem 14

Find the least odd prime factor of $2019^8+1$.

Solution 1

The problem tells us that $2019^8 \equiv -1 \pmod{p}$ for some prime $p$. We want to find the smallest odd possible value of $p$. By squaring both sides of the congruence, we get $2019^{16} \equiv 1 \pmod{p}$.

Since $2019^{16} \equiv 1 \pmod{p}$, $ord_p(2019)$ = $1, 2, 4, 8,$ or $16$

However, if $ord_p(2019)$ = $1, 2, 4,$ or $8,$ then $2019^8$ clearly will be $1 \pmod{p}$ instead of $-1 \pmod{p}$, causing a contradiction.

Therefore, $ord_p(2019) = 16$. Because $ord_p(2019)   \vert   \phi(p)$, $\phi(p)$ is a multiple of 16. Since we know $p$ is prime, $\phi(p) = p(1 - \frac{1}{p})$ or $p - 1$. Therefore, $p$ must be $1 \pmod{16}$. The two smallest primes that are $1 \pmod{16}$ are $17$ and $97$. $2019^8 \not\equiv -1 \pmod{17}$, but $2019^8 \equiv -1 \pmod{97}$, so our answer is $\boxed{097}$.

Note to solution 1

$\phi(p)$ is called the "Euler Function" of integer $p$. Eular theorem: define $\phi(p)$ as the number of positive integers less than $n$ but relatively prime to $n$, then we have \[\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})\] where $p_1,p_2,...,p_n$ are the prime factors of $p$. Then, we have \[a^{\phi(p)} \equiv 1\ (\mathrm{mod}\ p)\] if $(a,p)=1$.

Video Solution

On The Spot STEM:

https://youtu.be/_vHq5_5qCd8


https://youtu.be/IF88iO5keFo

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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