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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: Isn't it  <math>E</math>? <math>11^2 +26 = 147</math> which is divisible by <math>3</math>
 
Solution 2: Isn't it  <math>E</math>? <math>11^2 +26 = 147</math> which is divisible by <math>3</math>
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By: awesomemaths
  
 
==See Also==
 
==See Also==

Revision as of 16:27, 16 February 2018

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: Isn't it $E$? $11^2 +26 = 147$ which is divisible by $3$

By: awesomemaths

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AMC 10 Problems and Solutions

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