2019 AIME I Problems/Problem 14

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 $2019^{16} \equiv 1 \pmod{p}$ . This tells us that $\phi(p)$ 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}\(1-\frac{1}{p_i}\)\] (Error compiling LaTeX. Unknown error_msg)

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

2019 AIME I (Problems • Answer Key • Resources)
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2019 AIME I Problems/Problem 14

Contents

Problem 14

Solution 1

Note to solution 1

Video Solution

See Also