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

(Solution 2 (Answer Choices))
(Video Solution)
 
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~savannahsolver
 
~savannahsolver
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== Video Solution ==
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https://youtu.be/3bRjcrkd5mQ?t=187
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 03:20, 21 January 2021

Problem

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 $1 \pmod{3}$, the only expression is always a multiple of $3$ is $\boxed{\textbf{(C) } p^2+26}$. This is excluding when $p\equiv 0 \pmod{3}$, which only occurs when $p=3$, then $p^2+26=35$ which is still composite.

Solution 2 (Answer Choices)

Since the question asks which of the following will never be a prime number when $p^2$ is a prime number, a way to find the answer is by trying to find a value for $p$ such that the statement above won't be true.

A) $p^2+16$ isn't true when $p=5$ because $25+16=41$, which is prime

B) $p^2+24$ isn't true when $p=7$ because $49+24=73$, which is prime

C) $p^2+26$

D) $p^2+46$ isn't true when $p=5$ because $25+46=71$, which is prime

E) $p^2+96$ isn't true when $p=19$ because $361+96=457$, which is prime

Therefore, $\framebox{C}$ is the correct answer.

-DAWAE

Minor edit by Lucky1256. P=___ was the wrong number.

More minor edits by beanlol.

Video Solution

https://youtu.be/XRCWccGFnds

~savannahsolver

Video Solution

https://youtu.be/3bRjcrkd5mQ?t=187

~ pi_is_3.14

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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