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

Revision as of 16:48, 5 December 2020

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 28: / Line 28: @@
 Minor edit by Lucky1256. P=___ was the wrong number.
-Another miner edit by beanlol.
+More minor edits by beanlol.
 ==Video Solution==

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