Difference between revisions of "2019 AIME I Problems/Problem 14"
(→Note to solution 1) |
m |
||
| Line 6: | Line 6: | ||
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>. This tells us that <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>. | 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>. This tells us that <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=== |
<math>\phi(p)</math> is called the "Euler Function" of integer <math>p</math>. | <math>\phi(p)</math> is called the "Euler Function" of integer <math>p</math>. | ||
Eular 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>. | Eular 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>. | ||
Revision as of 18:08, 15 March 2019
Problem 14
Find the least odd prime factor of 
.
Solution 1
The problem tells us that 
for some prime 
. We want to find the smallest odd possible value of . By squaring both sides of the congruence, we get 
. This tells us that 
is a multiple of 16. Since we know is prime, 
or 
. Therefore, must be 
. The two smallest primes that are are 
and 
. 
, but 
, so our answer is 
.
Note to solution 1
is called the "Euler Function" of integer .
Eular theorem: define as the number of positive integers less than 
but relatively prime to , then we have ![\[\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})\]](http://latex.artofproblemsolving.com/3/7/9/3796cf787f69e13b3a5a060bcb1ba7d119e2150a.png)
where 
are the prime factors of . Then, we have ![\[a^{\phi(p)} \equiv 1\ (\mathrm{mod}\ p)\]](http://latex.artofproblemsolving.com/8/f/2/8f21dc4bc9438cd265fd03aadc87e1f0000e3808.png)
if 
.
Video Solution
On The Spot STEM:
See Also
| 2019 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
