Difference between revisions of "2018 AMC 10B Problems/Problem 11"
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=Solution= | =Solution= | ||
+ | ==Solution 1== | ||
+ | Each expression is in the form <math>p^2 + n</math>. | ||
+ | |||
+ | All prime numbers are of the form <math>6k \pm 1</math>, AKA they are congruent to <math>\pm1 \pmod{6}</math>. We can utilize this nicely to check for what we are looking for. If the expression is a prime, then | ||
+ | |||
+ | |||
+ | <math>p^2 + n \equiv \pm1 \pmod{6}</math> | ||
− | ==Solution | + | <math>\Rightarrow (\pm1)^2 + n \equiv \pm1 \pmod{6}</math> |
+ | |||
+ | <math>\Rightarrow 1 + n \equiv \pm1 \pmod{6}</math> | ||
+ | |||
+ | <math>\Rightarrow n \equiv (\pm1) - 1 \pmod{6}</math> | ||
+ | |||
+ | <math>\Rightarrow n \equiv (1</math> <math>or</math> <math>-1) - 1 \pmod{6}</math> | ||
+ | |||
+ | <math>\Rightarrow n \equiv 0</math> <math>or</math> <math>-2 \pmod{6}</math> | ||
+ | |||
+ | |||
+ | Now, just check for <math>n</math> in each option using this condition to check whether its prime or not. | ||
+ | |||
+ | <cmath>(A)\ n = 16; prime</cmath> | ||
+ | <cmath>(B)\ n = 24; prime</cmath> | ||
+ | <cmath>(C)\ n = 26; not\ prime</cmath> | ||
+ | <cmath>(D)\ n = 46; prime</cmath> | ||
+ | <cmath>(E)\ n = 96; prime</cmath> | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(C) } p^2 + 26} </math>. | ||
+ | |||
+ | ~ <math>shalomkeshet</math> | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
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. | 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 | + | |
+ | ==Solution 3 (Answer Choices)== | ||
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. | 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. | ||
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More minor edits by mathmonkey12. | More minor edits by mathmonkey12. | ||
+ | |||
==Video Solution (HOW TO THINK CREATIVELY!!!)== | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
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+ | ==Video Solution 1== | ||
+ | https://youtu.be/XRCWccGFnds | ||
− | |||
− | |||
− | == Video Solution == | + | == Video Solution 2== |
https://youtu.be/3bRjcrkd5mQ?t=187 | https://youtu.be/3bRjcrkd5mQ?t=187 | ||
+ | |||
==See Also== | ==See Also== |
Latest revision as of 11:41, 5 November 2024
Contents
[hide]Problem
Which of the following expressions is never a prime number when is a prime number?
Solution
Solution 1
Each expression is in the form .
All prime numbers are of the form , AKA they are congruent to . We can utilize this nicely to check for what we are looking for. If the expression is a prime, then
Now, just check for in each option using this condition to check whether its prime or not.
Therefore, the answer is .
~
Solution 2
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 3 (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 1
Video Solution 2
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.