Difference between revisions of "2019 AIME I Problems/Problem 14"
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==Problem 14== | ==Problem 14== | ||
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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| + | ==Solution== | ||
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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>. This tells us that <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>. | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 05:18, 15 March 2019
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 
. This tells us that 
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 
.
Video Solution
On The Spot STEM:
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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