Difference between revisions of "2019 AIME I Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
− | 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>. | + | 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>. |
− | + | Since <math>2019^{16} \equiv 1 \pmod{p}</math>, <math>ord_p(2019)</math> = <math>1, 2, 4, 8,</math> or <math>16</math> | |
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However, if <math>ord_p(2019)</math> = <math>1, 2, 4,</math> or <math>8,</math> then <math>2019^8</math> clearly will be <math>1 \pmod{p} </math> instead of <math>-1 \pmod{p}</math>, causing a contradiction. | However, if <math>ord_p(2019)</math> = <math>1, 2, 4,</math> or <math>8,</math> then <math>2019^8</math> clearly will be <math>1 \pmod{p} </math> instead of <math>-1 \pmod{p}</math>, causing a contradiction. | ||
− | Therefore, <math>ord_p(2019) = 16</math>, | + | Therefore, <math>ord_p(2019) = 16</math>. Because <math>ord_p(2019) \vert \phi(p)</math>, <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=== |
Revision as of 22:19, 16 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 .
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 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 .
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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.