Difference between revisions of "2018 AMC 10B Problems/Problem 11"
(→Solution 3) |
(→Solution 3) |
||
| Line 18: | Line 18: | ||
==Solution 3== | ==Solution 3== | ||
| − | From Fermat's Little Theorom, <math>p^2 \equiv 1\pmod 3</math> if <math>p</math> is | + | From Fermat's Little Theorom, <math>p^2 \equiv 1\pmod 3</math> if <math>p</math> is coprime with <math>3</math>. |
So for any <math>n\equiv2\pmod3</math>, <math>p^2+n \equiv 0\pmod 3</math> - divisible by 3, so not a prime. | So for any <math>n\equiv2\pmod3</math>, <math>p^2+n \equiv 0\pmod 3</math> - divisible by 3, so not a prime. | ||
The only choice <math>\equiv2\pmod3</math> is <math>\boxed{(\textbf{C})}</math> | The only choice <math>\equiv2\pmod3</math> is <math>\boxed{(\textbf{C})}</math> | ||
Revision as of 21:04, 22 February 2018
Which of the following expressions is never a prime number when 
is a prime number?
Solution 1
Because squares of a non-multiple of 3 is always 
, the only expression is always a multiple of 
is 
. This is excluding when 
, which only occurs when 
, then 
which is still composite.
Solution 2 (Bad Solution)
We proceed with guess and check:
.
Clearly only 
is our only option left.
(franchester)
Solution 3
From Fermat's Little Theorom, 
if is coprime with .
So for any 
, 
- divisible by 3, so not a prime.
The only choice 
is
Solution 4
Primes can only be 
or 
. Therefore, the square of a prime can only be 
. 
then must be 
, so it is always divisible by . Therefore, the answer is 
.
See Also
| 2018 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
