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Difference between revisions of "2018 AMC 10B Problems/Problem 11"

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==Solution 1==
 
==Solution 1==
  
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Because squares of a non-multiple of 3 is always <math>1 \pmod{3}</math>, the only expression is 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.
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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)==
  
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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.
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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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A) <math>p^2+16</math> isn't true when <math>p=5</math>
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A) <math>p^2+16</math> isn't true when <math>p=5</math> because <math>25+16=41</math>, which is prime
  
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B) <math>p^2+24</math> isn't true when <math>p=7</math>
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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>
  
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D) <math>p^2+46</math> isn't true when <math>p=5</math>
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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
  
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E) <math>p^2+96</math> isn't true when <math>p=19</math>  
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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.
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More minor edits by mathmonkey12.
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==Video Solution (HOW TO THINK CREATIVELY!!!)==
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https://youtu.be/PyCyMEBQCXM
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/XRCWccGFnds
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== Video Solution ==
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https://youtu.be/3bRjcrkd5mQ?t=187
  
 
==See Also==
 
==See Also==

Latest revision as of 13:35, 5 November 2023

Problem

Which of the following expressions is never a prime number when $p$ is a prime number?

$\textbf{(A) } p^2+16 \qquad \textbf{(B) } p^2+24 \qquad \textbf{(C) } p^2+26 \qquad \textbf{(D) } p^2+46 \qquad \textbf{(E) } p^2+96$

Solution 1

Because squares of a non-multiple of 3 is always $1 \pmod{3}$, the only expression always a multiple of $3$ is $\boxed{\textbf{(C) } p^2+26}$. This is excluding when $p\equiv 0 \pmod{3}$, which only occurs when $p=3$, then $p^2+26=35$ which is still composite.

Solution 2 (Answer Choices)

Since the question asks which of the following will never be a prime number when $p$ is a prime number, a way to find the answer is by trying to find a value for $p$ such that the statement above won't be true.

A) $p^2+16$ isn't true when $p=5$ because $25+16=41$, which is prime

B) $p^2+24$ isn't true when $p=7$ because $49+24=73$, which is prime

C) $p^2+26$

D) $p^2+46$ isn't true when $p=5$ because $25+46=71$, which is prime

E) $p^2+96$ isn't true when $p=19$ because $361+96=457$, which is prime

Therefore, $\framebox{C}$ 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!!!)

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

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

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