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Difference between revisions of "2018 AMC 10B Problems/Problem 11"
Line 10: Line 10:
  
 
We proceed with guess and check:
 
We proceed with guess and check:
−
<math>5^2+16=41</math>
+
<math>5^2+16=41 \qquad
−
<math>7^2+24=73</math>
+
7^2+24=73 \qquad
−
<math>11^2+26=147</math>
+
5^2+46=71 \qquad
−
<math>5^2+46=71</math>
+
19^2+96=457</math>.
−
Clearly only <math>\textbf{E}</math> is our only option left.
+
Clearly only <math>\boxed{(\textbf{C})}</math> is our only option left.
−
 
+
(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 (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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