Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution)
(→‎Solution)
 
(14 intermediate revisions by 10 users not shown)
Line 4: Line 4:
  
 
<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>
 +
 +
=Solution=
  
 
==Solution 1==
 
==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>
 +
  
−
Because squares of a non-multiple of 3 is always <math>1 \pmod{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\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 2==
  
−
==Solution 2 (Answer Choices)==
+
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.
  
−
Since the question asks which of the following will never be a prime number when <math>p^2</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.
+
 
 +
==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
 
A) <math>p^2+16</math> isn't true when <math>p=5</math> because <math>25+16=41</math>, which is prime
Line 31: Line 66:
 
More minor edits by beanlol.
 
More minor edits by beanlol.
  
−
==Video Solution==
+
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
 
https://youtu.be/XRCWccGFnds
  
−
~savannahsolver
 
  
−
== Video Solution ==
+
== Video Solution 2==
 
https://youtu.be/3bRjcrkd5mQ?t=187
 
https://youtu.be/3bRjcrkd5mQ?t=187
  
−
~ pi_is_3.14
 
  
 
==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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_11&oldid=231731"