Difference between revisions of "2018 AMC 10B Problems/Problem 11"
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D) <math>p^2+46</math> isn't true when <math>p=5</math> because <math>25+46=71</math>, which is prime | 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= | + | 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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More minor edits by beanlol. | More minor edits by beanlol. | ||
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+ | More minor edits by mathmonkey12. | ||
==Video Solution (HOW TO THINK CREATIVELY!!!)== | ==Video Solution (HOW TO THINK CREATIVELY!!!)== |
Revision as of 13:35, 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.