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

Revision as of 19:41, 25 September 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 $p$ . Euler theorem: define $\phili(p)$ (Error compiling LaTeX. Unknown error_msg) 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 $(a,p)=1$ .

Furthermore, $ord_n(a)$ for an integer $a$ relatively prime to $n$ is defined as the smallest positive integer $d$ such that $a^{d} \equiv 1\ (\mathrm{mod}\ n)$ . An important property of the order is that $ord_n(a)|\phi(n)$ .

Video Solution

On The Spot STEM:

https://youtu.be/_vHq5_5qCd8

https://youtu.be/IF88iO5keFo

@@ Line 14: / Line 14: @@
 ===Note to solution 1===
 <math>\phi(p)</math> is called the "Euler Function" of integer <math>p</math>.
-Euler theorem: define <math>\phi(p)</math> as the number of positive integers less than <math>n</math> but relatively prime to <math>n</math>, then we have <cmath>\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})</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\ (\mathrm{mod}\ p)</cmath> if <math>(a,p)=1</math>.
+Euler theorem: define <math>\phili(p)</math> as the number of positive integers less than <math>n</math> but relatively prime to <math>n</math>, then we have <cmath>\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})</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\ (\mathrm{mod}\ p)</cmath> if <math>(a,p)=1</math>.
 Furthermore, <math>ord_n(a)</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\ (\mathrm{mod}\ n)</math>. An important property of the order is that <math>ord_n(a)|\phi(n)</math>.

2019 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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