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

(Solution 2 (Bad Solution))
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5^2+46=71 \qquad
 
5^2+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)
 
(franchester)
 +
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2018|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2018|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:17, 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 (Bad Solution)

We proceed with guess and check: $5^2+16=41 \qquad 7^2+24=73 \qquad 5^2+46=71 \qquad 19^2+96=457$. Clearly only $\boxed{(\textbf{C})}$ is our only option left. (franchester)

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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