Difference between revisions of "2018 AMC 10B Problems/Problem 11"
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==Solution 2 (Answer Choices)== | ==Solution 2 (Answer Choices)== | ||
− | Since the question asks which of the following will never be a prime number when p^2 is a prime number, a way to find the answer is by trying to find a value for <math>p</math> such that the statement above won't be true. | + | Since the question asks which of the following will never be a prime number when <math>p^2</math> is a prime number, a way to find the answer is by trying to find a value for <math>p</math> such that the statement above won't be true. |
− | A) p^2+16 isn't true when p=5 | + | A) <math>p^2+16</math> isn't true when <math>p=5</math> |
− | B) p^2+24 isn't true when p=7 | + | B) <math>p^2+24</math> isn't true when p=7 |
− | C) | + | C) <math>p^2+26</math> |
− | D) p^2+46 isn't true when p=11 | + | D) <math>p^2+46</math> isn't true when p=11 |
− | E) p^2+96 isn't true when p=17. | + | E) <math>p^2+96</math> isn't true when p=17. |
Therefore, <math>C</math> is the correct answer. | Therefore, <math>C</math> is the correct answer. | ||
Revision as of 21:34, 2 September 2019
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 (Answer Choices)
Since the question asks which of the following will never be a prime number when is a prime number, a way to find the answer is by trying to find a value for
such that the statement above won't be true.
A)
isn't true when
B)
isn't true when p=7
C)
D)
isn't true when p=11
E)
isn't true when p=17.
Therefore,
is the correct answer.
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.