Difference between revisions of "2019 AIME I Problems/Problem 14"

Revision as of 20:27, 14 March 2019

The 2019 AIME I takes place on March 13, 2019.

STOP IT!

CUT IT OUT.

@@ Line 5: / Line 5: @@
 ==Solution 2==
-Essentially, we are trying to find the smallest prime p such that <math>2019^8 \equiv -1 (\mod p)</math>. This congruence tells us that <math>2019^{16} \equiv 1 (\mod p)</math>. 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 <math>\phi(p)</math>, which is just p-1 because p is prime. Thus, we have <math>16|p-1</math>. Now, we just test up. 17 does not work, because <math>2019^8 +1</math> reduces to 2 modulo 17. The reason this does not work is that <math>2019^8</math> 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
+CUT IT OUT.
-<math>2019^8 \equiv (-18)^8 \equiv 324^4 \equiv 33^4 \equiv 1089^2 \equiv 22^2 \equiv 96 (\mod 97)</math>.
-Thus <math>2019^8 +1 \equiv 0 (\mod\boxed{97})</math>.-vvluo
-also not the most rigorous(last part)
 ==See Also==
 {{AIME box|year=2019|n=I|num-b=13|num-a=15}}
 {{MAA Notice}}

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