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

Revision as of 03:25, 21 January 2023

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 $p$ 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 by OmegaLearn

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

~ pi_is_3.14

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

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

@@ Line 36: / Line 36: @@
 ~savannahsolver
-== Video Solution ==
+== Video Solution by OmegaLearn==
 https://youtu.be/3bRjcrkd5mQ?t=187

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