Difference between revisions of "2019 AIME I Problems/Problem 14"
(→Video Solution) |
m (→Note to solution) |
||
Line 14: | Line 14: | ||
===Note to solution=== | ===Note to solution=== | ||
<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>. | ||
− | Euler theorem: define <math>\phi(p)</math> as the number of positive integers less than <math>p</math> but relatively prime to <math>p</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>\phi(p)</math> as the number of positive integers less than <math>p</math> but relatively prime to <math>p</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>\gcd(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>. | 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>. |
Revision as of 18:50, 16 March 2020
Contents
Problem 14
Find the least odd prime factor of .
Solution
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 .
Since , = or
However, if = or then clearly will be instead of , causing a contradiction.
Therefore, . Because , 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
is called the "Euler Function" of integer . Euler theorem: define as the number of positive integers less than but relatively prime to , then we have where are the prime factors of . Then, we have if .
Furthermore, for an integer relatively prime to is defined as the smallest positive integer such that . An important property of the order is that .
Video Solution
On The Spot STEM:
Solution 2(basbhbashabsbhbasbasbhasbhsabhasbhbhasbhbash)
1. Take remainder of prime
2. Take that to the power of , mod
3. If it is congruent to 1, you are done
4. If not, repeat for next prime
5. Through 2 hours and 59 minutes of bashing, we arrive at the answer of -Trex4days
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.