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"

(This solution is incorrect)
m (→‎Solution 2)
Line 7: Line 7:
 
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\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.
  
−
==Solution 3==
+
==Solution 2==
 
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>.
 
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>.
  

Revision as of 04:51, 20 January 2019

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 $1\mod 3$, the only expression is always a multiple of $3$ is $\boxed{\textbf{(C) } p^2+26}$. This is excluding when $p=0\mod3$, which only occurs when $p=3$, then $p^2+26=35$ which is still composite.

Solution 2

Primes can only be $1$ or $-1\mod 6$. Therefore, the square of a prime can only be $1\mod 6$. $p^2+26$ then must be $3\mod 6$, so it is always divisible by $3$. Therefore, the answer is $\boxed{\text{(C)}}$.

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=100668"