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

Revision as of 09:42, 8 October 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 $\boxed{97}$ .

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 10: / Line 10: @@
 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>. 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>.
+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{97}</math>.
 ===Note to solution 1===

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