Difference between revisions of "2019 AIME I Problems/Problem 14"
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==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} | + | 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>. |
==Video Solution== | ==Video Solution== |
Revision as of 09:35, 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 where are the prime factors of . Then, we have 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 |
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