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

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==Problem==
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Which of the following expressions is never a prime number when <math>p</math> is a prime number?
 
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>
 
<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>
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=Solution=
  
 
==Solution 1==
 
==Solution 1==
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Each expression is in the form <math>p^2 + n</math>.
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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
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<math>p^2 + n \equiv \pm1 \pmod{6}</math>
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<math>\Rightarrow (\pm1)^2 + n \equiv \pm1 \pmod{6}</math>
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<math>\Rightarrow 1 + n \equiv \pm1 \pmod{6}</math>
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<math>\Rightarrow n \equiv (\pm1) - 1 \pmod{6}</math>
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<math>\Rightarrow n \equiv (1</math> <math>or</math> <math>-1) - 1 \pmod{6}</math>
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<math>\Rightarrow n \equiv 0</math> <math>or</math> <math>-2 \pmod{6}</math>
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Now, just check for <math>n</math> in each option using this condition to check whether its prime or not.
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<cmath>(A)\ n = 16; prime</cmath>
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<cmath>(B)\ n = 24; prime</cmath>
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<cmath>(C)\ n = 26; not\ prime</cmath>
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<cmath>(D)\ n = 46; prime</cmath>
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<cmath>(E)\ n = 96; prime</cmath>
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Therefore, the answer is <math>\boxed{\textbf{(C) } p^2 + 26} </math>.
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~ <math>shalomkeshet</math>
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==Solution 2==
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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.
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==Solution 3 (Answer Choices)==
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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.
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A) <math>p^2+16</math> isn't true when <math>p=5</math> because <math>25+16=41</math>, which is prime
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B) <math>p^2+24</math> isn't true when <math>p=7</math> because <math>49+24=73</math>, which is prime
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C) <math>p^2+26</math>
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D) <math>p^2+46</math> isn't true when <math>p=5</math> because <math>25+46=71</math>, which is prime
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E) <math>p^2+96</math> isn't true when <math>p=19</math> because <math>361+96=457</math>, which is prime
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Therefore, <math>\framebox{C}</math> is the correct answer.
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-DAWAE
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Minor edit by Lucky1256. P=___ was the wrong number.
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More minor edits by beanlol.
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More minor edits by mathmonkey12.
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==Video Solution (HOW TO THINK CREATIVELY!!!)==
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https://youtu.be/PyCyMEBQCXM
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~Education, the Study of Everything
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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\mod3</math>, which only occurs when <math>p=3</math>, then <math>p^2+26=35</math> which is still composite.
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==Video Solution 1==
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https://youtu.be/XRCWccGFnds
  
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==Solution 2 (Not Recommended Solution)==
 
  
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We proceed with guess and check:
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== Video Solution 2==
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<math>5^2+16=41 \qquad
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https://youtu.be/3bRjcrkd5mQ?t=187
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7^2+24=73 \qquad
 
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5^2+46=71 \qquad
 
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19^2+96=457</math>.
 
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Clearly only <math>\boxed{(\textbf{C})}</math> is our only option left.
 
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(franchester)
 
  
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==Solution 3==
 
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Primes can only be <math>1</math> or <math>-1\mod 6</math>. Therefore, the square of a prime can only be <math>1\mod 6</math>. <math>p^2+26</math> then must be <math>3\mod 6</math>, so it is always divisible by <math>3</math>. Therefore, the answer is <math>\boxed{\text{(C)}}</math>.
 
  
 
==See Also==
 
==See Also==

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


See Also

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

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