Difference between revisions of "2018 AMC 10B Problems/Problem 11"
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==Solution 1== | ==Solution 1== | ||
− | Because squares of a non-multiple of 3 is always <math>1 \pmod{3}</math>, the only expression | + | Because squares of a non-multiple of 3 is always <math>1 \pmod{3}</math>, the only expression always a multiple of <math>3</math> is <math>\boxed{\textbf{(C) } p^2+26} </math>. This is excluding when <math>p\equiv 0 \pmod{3}</math>, which only occurs when <math>p=3</math>, then <math>p^2+26=35</math> which is still composite. |
==Solution 2 (Answer Choices)== | ==Solution 2 (Answer Choices)== | ||
− | Since the question asks which of the following will never be a prime number when <math>p | + | Since the question asks which of the following will never be a prime number when <math>p</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) <math>p^2+16</math> isn't true when <math>p=5</math> | + | A) <math>p^2+16</math> isn't true when <math>p=5</math> because <math>25+16=41</math>, which is prime |
− | B) <math>p^2+24</math> isn't true when <math>p=7</math> | + | B) <math>p^2+24</math> isn't true when <math>p=7</math> because <math>49+24=73</math>, which is prime |
C) <math>p^2+26</math> | C) <math>p^2+26</math> | ||
− | D) <math>p^2+46</math> isn't true when <math>p=5</math> | + | D) <math>p^2+46</math> isn't true when <math>p=5</math> because <math>25+46=71</math>, which is prime |
− | E) <math>p^2+96</math> isn't true when <math>p=19</math> | + | E) <math>p^2+96</math> isn't true when <math>p=19</math> because <math>361+96=457</math>, which is prime |
Therefore, <math>\framebox{C}</math> is the correct answer. | Therefore, <math>\framebox{C}</math> is the correct answer. | ||
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Minor edit by Lucky1256. P=___ was the wrong number. | Minor edit by Lucky1256. P=___ was the wrong number. | ||
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+ | More minor edits by beanlol. | ||
+ | |||
+ | More minor edits by mathmonkey12. | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/PyCyMEBQCXM | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/XRCWccGFnds | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/3bRjcrkd5mQ?t=187 | ||
==See Also== | ==See Also== |
Latest revision as of 14:35, 5 November 2023
Contents
Problem
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 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
because
, which is prime
B) isn't true when
because
, which is prime
C)
D) isn't true when
because
, which is prime
E) isn't true when
because
, which is prime
Therefore, is the correct answer.
-DAWAE
Minor edit by Lucky1256. P=___ was the wrong number.
More minor edits by beanlol.
More minor edits by mathmonkey12.
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution
Video Solution
https://youtu.be/3bRjcrkd5mQ?t=187
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.