Revision as of 14:41, 2 September 2025

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

Solution 1

Each expression is in the form $p^2 + n$ .

All prime numbers are of the form $6k \pm 1$ , AKA they are congruent to $\pm1 \pmod{6}$ . We can utilize this nicely to check for what we are looking for. If the expression is a prime, then

$p^2 + n \equiv \pm1 \pmod{6}$

$\Rightarrow (\pm1)^2 + n \equiv \pm1 \pmod{6}$

$\Rightarrow 1 + n \equiv \pm1 \pmod{6}$

$\Rightarrow n \equiv (\pm1) - 1 \pmod{6}$

$\Rightarrow n \equiv (1$ $or$ $-1) - 1 \pmod{6}$

$\Rightarrow n \equiv 0$ $or$ $-2 \pmod{6}$

Now, just check for $n$ in each option using this condition to check whether its prime or not.

$(A)\ n = 16; prime$ $(B)\ n = 24; prime$ $(C)\ n = 26; not\ prime$ $(D)\ n = 46; prime$ $(E)\ n = 96; prime$

Therefore, the answer is $\boxed{\textbf{(C) } p^2 + 26}$ .

~ $shalomkeshet$

Solution 2

Because squares of a non-multiple of 3 is always $1 \pmod{3}$ , the only expression 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 3 (Solution 4 but better)

$ p^2 + 16 $ is prime for $ p = 5 $, so we know that $ A $ doesn't work.

Because $ p^2 + 96 = p^2 + (16) \cdot 6 $, we see that eventually there exists a number $ p $ that makes the sum prime, therefore $ E $ also doesn't work.

Option $ B $ is $ p^2 + 24 $, but because the number itself has no carryover, we can assume $ p = 5 $ again to get the number $ 41 $, which is prime.

Option $ D $ is $ p^2 + 46 $, ending in $ 6 $, so we try $ p = 5 $ again to get $ 71 $. This is also prime.

We can either deduce through elimination that $ p^2 + 26 $ is always composite for any prime $ p $, but we can prove it by showing $ p = 5 $ doesn't work, $ p = 7 $ doesn't work, and $ p = 11 $ doesn't work, and through Engineer's Induction, $ p^2 + 26 $ always remains composite.

Therefore the answer is $\boxed{\textbf{(C) } p^2 + 26}$ .

~Pinotation

Solution 4 (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.

More minor edits by mathmonkey12.

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/PyCyMEBQCXM

~Education, the Study of Everything

Video Solution 1

https://youtu.be/XRCWccGFnds

Video Solution 2

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

@@ Line 53: / Line 53: @@
 Option \( D \) is \( p^2 + 46 \), ending in \( 6 \), so we try \( p = 5 \) again to get \( 71 \). This is also prime.
-We can either deduce through elimination that \( p^2 + 26 \) is always composite for any prime \( p \), but we can prove it by showing \( p = 5 \) doesn't work, \( p = 7 \) doesn't work, and \( p = 11 \) doesn't work, and through Engineer's Induction, \( p^2 + 26 \) always remains composite.
+We can either deduce through elimination that \( p^2 + 26 \) is always composite for any prime \( p \), but we can prove it by showing \( p = 5 \) doesn't work, \( p = 7 \) doesn't work, and \( p = 11 \) doesn't work, and through [[Engineer's Induction]], \( p^2 + 26 \) always remains composite.
 Therefore the answer is <math>\boxed{\textbf{(C) } p^2 + 26} </math>.
+~Pinotation
 ==Solution 4 (Answer Choices)==

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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