2018 AMC 10B Problems/Problem 11

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)

Solution 3

From Fermat's Little Theorom, $p^2 \equiv 1\pmod 3$ if $p$ is co-prime with $3$ . So for any $n\equiv2\pmod3$ , $p^2+n \equiv 0\pmod 3$ - divisible by 3, so not a prime. The only choice $\equiv2\pmod3$ is $\boxed{(\textbf{C})}$

Solution 4

Primes can only be $1$ or $-1\mod 6$ . Therefore, the square of a prime can only be $1\mod 6$ . $p^2+26$ then must be $3\mod 6$ , so it is always divisible by $3$ . Therefore, the answer is $\boxed{\text{(C)}}$ .

2018 AMC 10B (Problems • Answer Key • Resources)
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2018 AMC 10B Problems/Problem 11

Contents

Solution 1

Solution 2 (Bad Solution)

Solution 3

Solution 4

See Also