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

Revision as of 20:41, 2 September 2019

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 (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 $p=5$

B) $p^2+24$ isn't true when $p=7$

C) $p^2+26$

D) $p^2+46$ isn't true when $p=11$

E) $p^2+96$ isn't true when $p=17$

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

-DAWAE

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.

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 Therefore, <math>\framebox{C}</math> is the correct answer.
+-DAWAE
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Difference between revisions of "2018 AMC 10B Problems/Problem 11"

Revision as of 20:41, 2 September 2019

Solution 1

Solution 2 (Answer Choices)

See Also