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

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Because squares of a non-multiple of 3 is always <math>1\mod 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=0\mod3</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\mod 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=0\mod3</math>, which only occurs when <math>p=3</math>, then <math>p^2+26=35</math> which is still composite.
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==Solution 2 (Answer Choices)==
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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.
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A) p^2+16 isn't true when p=5
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B) p^2+24 isn't true when p=7
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C) P^2+26
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D) p^2+46 isn't true when p=11
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E) p^2+96 isn't true when p=17.
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Therefore, <math>C</math> is the correct answer.
  
 
==See Also==
 
==See Also==

Revision as of 20:48, 22 August 2019

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 $1\mod 3$, the only expression is always a multiple of $3$ is $\boxed{\textbf{(C) } p^2+26}$. This is excluding when $p=0\mod3$, 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^2 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 B) p^2+24 isn't true when p=7 C) P^2+26 D) p^2+46 isn't true when p=11 E) p^2+96 isn't true when p=17. Therefore, $C$ 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

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