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"
m (→‎Solution 2 (Answer Choices))
(50 intermediate revisions by 28 users not shown)
Line 1: Line 1:
 +
==Problem==
 +
 
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?
  
Line 5: Line 7:
 
==Solution 1==
 
==Solution 1==
  
−
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.
+
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 2 (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
 +
 
 +
 
  
−
==Solution 2 (Bad Solution)==
+
==Video Solution==
 +
https://youtu.be/XRCWccGFnds
  
−
We proceed with guess and check:
+
== Video Solution ==
−
<math>5^2+16=41</math>
+
https://youtu.be/3bRjcrkd5mQ?t=187
−
<math>7^2+24=73</math>
 
−
<math>11^2+26=147</math>
 
−
<math>5^2+46=71</math>
 
−
Clearly only <math>\textbf{E}</math> is our only option left.
 
  
 
==See Also==
 
==See Also==

Revision as of 13:35, 5 November 2023

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 1

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 2 (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

https://youtu.be/XRCWccGFnds

Video Solution

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=200847"