Difference between revisions of "2019 AIME I Problems/Problem 14"
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==Solution== | ==Solution== | ||
− | + | We know 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 find <math>2019^{16} \equiv 1 \pmod{p}</math>. | |
− | Since <math>2019^{16} \equiv 1 \pmod{p}</math>, <math> | + | Since <math>2019^{16} \equiv 1 \pmod{p}</math>, the order of <math>2019</math> modulo <math>p</math> is <math>1, 2, 4, 8,</math> or <math>16</math>. |
− | However, if <math> | + | However, if the order of <math>2019</math> modulo <math>p</math> is <math>1, 2, 4,</math> or <math>8,</math> then <math>2019^8</math> will be equivalent to <math>1 \pmod{p},</math> which contradicts the given requirement that <math>2019^8\equiv -1\pmod{p}</math>. |
− | Therefore, <math> | + | Therefore, the order of <math>2019</math> modulo <math>p</math> is <math>16</math>. Because all orders modulo <math>p</math> divide <math>\phi(p)</math>, we see that <math>\phi(p)</math> is a multiple of 16. As <math>p</math> is prime, <math>\phi(p) = p(1 - \frac{1}{p}) = p - 1</math>. Therefore, <math>p\equiv1 \pmod{16}</math>. The two smallest primes equivalent to <math>1 \pmod{16}</math> are <math>17</math> and <math>97</math>. As <math>2019^8 \not\equiv -1 \pmod{17}</math> and <math>2019^8 \equiv -1 \pmod{97}</math>, our answer is <math>\boxed{97}</math>. |
===Note to solution=== | ===Note to solution=== | ||
− | <math>\phi( | + | <math>\phi(k)</math> is the [[Euler Totient Function]] of integer <math>k</math>. |
[[Euler's Totient 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>. | [[Euler's Totient 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> | + | Furthermore, the order <math>a</math> modulo <math>n</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\pmod n</math>. An important property of the order <math>d</math> is that <math>d|\phi(n)</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 19:09, 18 May 2020
Problem 14
Find the least odd prime factor of .
Solution
We know that for some prime . We want to find the smallest odd possible value of . By squaring both sides of the congruence, we find .
Since , the order of modulo is or .
However, if the order of modulo is or then will be equivalent to which contradicts the given requirement that .
Therefore, the order of modulo is . Because all orders modulo divide , we see that is a multiple of 16. As is prime, . Therefore, . The two smallest primes equivalent to are and . As and , our answer is .
Note to solution
is the Euler Totient Function of integer . Euler's Totient 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, the order modulo 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:
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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