Latest revision as of 11:41, 5 November 2024

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 (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 1: / Line 1: @@
+==Problem==
 Which of the following expressions is never a prime number when <math>p</math> is a prime number?
 <math>\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</math>
+=Solution=
 ==Solution 1==
+Each expression is in the form <math>p^2 + n</math>.
+All prime numbers are of the form <math>6k \pm 1</math>, AKA they are congruent to <math>\pm1 \pmod{6}</math>. We can utilize this nicely to check for what we are looking for. If the expression is a prime, then
+<math>p^2 + n \equiv \pm1 \pmod{6}</math>
+<math>\Rightarrow (\pm1)^2 + n \equiv \pm1 \pmod{6}</math>
+<math>\Rightarrow 1 + n \equiv \pm1 \pmod{6}</math>
+<math>\Rightarrow n \equiv (\pm1) - 1 \pmod{6}</math>
+<math>\Rightarrow n \equiv (1</math> <math>or</math> <math>-1) - 1 \pmod{6}</math>
+<math>\Rightarrow n \equiv 0</math> <math>or</math> <math>-2 \pmod{6}</math>
+Now, just check for <math>n</math> in each option using this condition to check whether its prime or not.
+<cmath>(A)\ n = 16; prime</cmath>
+<cmath>(B)\ n = 24; prime</cmath>
+<cmath>(C)\ n = 26; not\ prime</cmath>
+<cmath>(D)\ n = 46; prime</cmath>
+<cmath>(E)\ n = 96; prime</cmath>
+Therefore, the answer is <math>\boxed{\textbf{(C) } p^2 + 26} </math>.
+~ <math>shalomkeshet</math>
+==Solution 2==
+Because squares of a non-multiple of 3 is always <math>1 \pmod{3}</math>, the only expression always a multiple of <math>3</math> is <math>\boxed{\textbf{(C) } p^2+26} </math>. This is excluding when <math>p\equiv 0 \pmod{3}</math>, which only occurs when <math>p=3</math>, then <math>p^2+26=35</math> which is still composite.
+==Solution 3 (Answer Choices)==
+Since the question asks which of the following will never be a prime number when <math>p</math> is a prime number, a way to find the answer is by trying to find a value for <math>p</math> such that the statement above won't be true.
+A) <math>p^2+16</math> isn't true when <math>p=5</math> because <math>25+16=41</math>, which is prime
+B) <math>p^2+24</math> isn't true when <math>p=7</math> because <math>49+24=73</math>, which is prime
+C) <math>p^2+26</math>
+D) <math>p^2+46</math> isn't true when <math>p=5</math> because <math>25+46=71</math>, which is prime
+E) <math>p^2+96</math> isn't true when <math>p=19</math> because <math>361+96=457</math>, which is prime
+Therefore, <math>\framebox{C}</math> 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
+==See Also==
-Because squares of a non-multiple of 3 is always <math>1\mod 3</math>, the only expression is always a multiple of <math>3</math> is <math>\boxed{\textbf{(C) } p^2+26} </math>. This is excluding when <math>p=0\mod 3</math>, then <math>p^2+26=35</math> which is still composite.
+{{AMC10 box|year=2018|ab=B|num-b=10|num-a=12}}
+{{MAA Notice}}

2018 AMC 10B (Problems • Answer Key • Resources)
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