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.
  
==Solution 3==
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==Solution 2==
 
Primes can only be <math>1</math> or <math>-1\mod 6</math>. Therefore, the square of a prime can only be <math>1\mod 6</math>. <math>p^2+26</math> then must be <math>3\mod 6</math>, so it is always divisible by <math>3</math>. Therefore, the answer is <math>\boxed{\text{(C)}}</math>.
 
Primes can only be <math>1</math> or <math>-1\mod 6</math>. Therefore, the square of a prime can only be <math>1\mod 6</math>. <math>p^2+26</math> then must be <math>3\mod 6</math>, so it is always divisible by <math>3</math>. Therefore, the answer is <math>\boxed{\text{(C)}}</math>.
  

Revision as of 03:51, 20 January 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

Primes can only be $1$ or $-1\mod 6$. Therefore, the square of a prime can only be $1\mod 6$. $p^2+26$ then must be $3\mod 6$, so it is always divisible by $3$. Therefore, the answer is $\boxed{\text{(C)}}$.

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 10 Problems and Solutions

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