Difference between revisions of "2019 AIME I Problems/Problem 14"
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+ | ==Solution 2(basbhbashabsbhbasbasbhasbhsabhasbhbhasbhbash)== | ||
+ | 1. Take remainder of prime <math>p</math> | ||
+ | |||
+ | 2. Take that to the power of <math>8</math>, mod <math>p</math> | ||
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+ | 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 <math>\boxed{097}</math> | ||
+ | -Trex4days | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=13|num-a=15}} | {{AIME box|year=2019|n=I|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:40, 4 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.