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

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(Solution 2 (Not Recommended Solution))
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We proceed with guess and check:
 
We proceed with guess and check:
<math>5^2+16=41 \qquad
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<math>3^2+16=25 \qquad
7^2+24=73 \qquad
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1^2+24=25 \qquad
5^2+46=71 \qquad
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1^2+46=71 \qquad
 
19^2+96=457</math>.
 
19^2+96=457</math>.
 
Clearly only <math>\boxed{(\textbf{C})}</math> is our only option left.
 
Clearly only <math>\boxed{(\textbf{C})}</math> is our only option left.
(franchester)
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-liu4505
  
 
==Solution 3==
 
==Solution 3==

Revision as of 00:29, 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 (Not Recommended Solution)

We proceed with guess and check: $3^2+16=25 \qquad 1^2+24=25 \qquad 1^2+46=71 \qquad 19^2+96=457$. Clearly only $\boxed{(\textbf{C})}$ is our only option left. -liu4505

Solution 3

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
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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