Difference between revisions of "2019 AIME I Problems/Problem 14"
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Find the least odd prime factor of <math>2019^8+1</math>. | Find the least odd prime factor of <math>2019^8+1</math>. | ||
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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>. |
Revision as of 14:41, 25 November 2019
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 1
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
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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 |
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