Difference between revisions of "2018 AMC 10B Problems/Problem 11"
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Primes can only be <math>1</math> or <math>-1\mod 6</math>. Therefore, the square of a prime can only be <math>1\mod 6</math>. <math>p^2+26</math> then must be <math>3\mod 6</math>, so it is always divisible by <math>3</math>. Therefore, the answer is <math>\boxed{\text{(C)}}</math>. | Primes can only be <math>1</math> or <math>-1\mod 6</math>. Therefore, the square of a prime can only be <math>1\mod 6</math>. <math>p^2+26</math> then must be <math>3\mod 6</math>, so it is always divisible by <math>3</math>. Therefore, the answer is <math>\boxed{\text{(C)}}</math>. | ||
Revision as of 20:59, 3 September 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 (Not Recommended Solution)
We proceed with guess and check:
.
Clearly only
is our only option left.
(franchester)
Solution 3
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.