Difference between revisions of "2019 AIME I Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
Revision as of 19:27, 14 March 2019
The 2019 AIME I takes place on March 13, 2019.
Problem 14
STOP IT!
Solution 2
Essentially, we are trying to find the smallest prime p such that 
. This congruence tells us that 
. Therefore, the order of 2019 modulo p is a divisor of 16, but not a divisor of 8. This tells us that the order of 2019 modulo p is exactly 16 since it is the only possibility. We know that the order of 2019 modulo p is a divisor of 
, which is just p-1 because p is prime. Thus, we have 
. Now, we just test up. 17 does not work, because 
reduces to 2 modulo 17. The reason this does not work is that 
reduces to 1, not -1 modulo 17. Since, 33, 49, 65, and 81 are all composite, we are able to skip those cases. This brings us to 97, which gives
.
Thus 
.-vvluo
also not the most rigorous(last part)
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. 
