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

Revision as of 19:22, 18 May 2020

Problem 14

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

Solution

We know 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 find $2019^{16} \equiv 1 \pmod{p}$ .

Since $2019^{16} \equiv 1 \pmod{p}$ , the order of $2019$ modulo $p$ is a positive divisor of $16$ .

However, if the order of $2019$ modulo $p$ is $1, 2, 4,$ or $8,$ then $2019^8$ will be equivalent to $1 \pmod{p},$ which contradicts the given requirement that $2019^8\equiv -1\pmod{p}$ .

Therefore, the order of $2019$ modulo $p$ is $16$ . Because all orders modulo $p$ divide $\phi(p)$ , we see that $\phi(p)$ is a multiple of $16$ . As $p$ is prime, $\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1$ . Therefore, $p\equiv 1 \pmod{16}$ . The two smallest primes equivalent to $1 \pmod{16}$ are $17$ and $97$ . As $2019^8 \not\equiv -1 \pmod{17}$ and $2019^8 \equiv -1 \pmod{97}$ , the smallest possible $p$ is thus $\boxed{097}$ .

Note to solution

$\phi(k)$ is the Euler Totient Function of integer $k$ . $\phi(k)$ is the number of positive integers less than $k$ relatively prime to $k$ . Define the numbers $k_1,k_2,k_3,\cdots,k_n$ to be the prime factors of $k$ . Then, we have $\phi(k)=k\cdot \prod^n_{i=1}\left(1-\dfrac{1}{k_i}\right).$ $A property of the Totient function is that, for any prime$ p $,$ \phi(p)=p-1$.

[[Euler's Totient Theorem]] states that <cmath>a^{\phi(k)} \equiv 1\pmod k</cmath> if$ (Error compiling LaTeX. Unknown error_msg)\gcd(a,k)=1$.

Furthermore, the order$ (Error compiling LaTeX. Unknown error_msg)a $modulo$ n $for an integer$ a $relatively prime to$ n $is defined as the smallest positive integer$ d $such that$ a^{d} \equiv 1\pmod n $. An important property of the order$ d $is that$ d|\phi(n)$.

Video Solution

On The Spot STEM:

https://youtu.be/_vHq5_5qCd8

https://youtu.be/IF88iO5keFo

@@ Line 13: / Line 13: @@
 ===Note to solution===
-<math>\phi(p)</math> is the [[Euler Totient Function]] of integer <math>p</math>.
+<math>\phi(k)</math> is the [[Euler Totient Function]] of integer <math>k</math>. <math>\phi(k)</math> is the number of positive integers less than <math>k</math> relatively prime to <math>k</math>. Define the numbers <math>k_1,k_2,k_3,\cdots,k_n</math> to be the prime factors of <math>k</math>. Then, we have <math>\phi(k)=k\cdot \prod^n_{i=1}\left(1-\dfrac{1}{k_i}\right).</math><math> A property of the Totient function is that, for any prime </math>p<math>, </math>\phi(p)=p-1<math>.
-[[Euler's Totient Theorem]]: define <math>\phi(p)</math> as the number of positive integers less than <math>p</math> but relatively prime to <math>p</math>. We have <cmath>\phi(p)=p\cdot \prod^n_{i=1}\left(1-\dfrac{1}{p_i}\right)</cmath> where <math>p_1,p_2,...,p_n</math> are the prime factors of <math>p</math>. Then, we have <cmath>a^{\phi(p)} \equiv 1\pmod p</cmath> if <math>\gcd(a,p)=1</math>.
-Furthermore, the order <math>a</math> modulo <math>n</math> for an integer <math>a</math> relatively prime to <math>n</math> is defined as the smallest positive integer <math>d</math> such that <math>a^{d} \equiv 1\pmod n</math>. An important property of the order <math>d</math> is that <math>d|\phi(n)</math>.
+[[Euler's Totient Theorem]] states that <cmath>a^{\phi(k)} \equiv 1\pmod k</cmath> if </math>\gcd(a,k)=1<math>.
+Furthermore, the order </math>a<math> modulo </math>n<math> for an integer </math>a<math> relatively prime to </math>n<math> is defined as the smallest positive integer </math>d<math> such that </math>a^{d} \equiv 1\pmod n<math>. An important property of the order </math>d<math> is that </math>d|\phi(n)$.
 ==Video Solution==

2019 AIME I (Problems • Answer Key • Resources)
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