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"

(Added 4th solution)
(→‎Solution 3)
Line 18: Line 18:
  
 
==Solution 3==
 
==Solution 3==
−
From Fermat's Little Theorom, <math>p^2 \equiv 1\pmod 3</math> if <math>p \neq 3</math>.
+
From Fermat's Little Theorom, <math>p^2 \equiv 1\pmod 3</math> if <math>p</math> is co-prime with <math>3</math>.
 
So for any <math>n\equiv2\pmod3</math>, <math>p^2+n \equiv 0\pmod 3</math> - divisible by 3, so not a prime.  
 
So for any <math>n\equiv2\pmod3</math>, <math>p^2+n \equiv 0\pmod 3</math> - divisible by 3, so not a prime.  
 
The only choice <math>\equiv2\pmod3</math> is <math>\boxed{(\textbf{C})}</math>
 
The only choice <math>\equiv2\pmod3</math> is <math>\boxed{(\textbf{C})}</math>

Revision as of 21:04, 22 February 2018

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 (Bad Solution)

We proceed with guess and check: $5^2+16=41 \qquad 7^2+24=73 \qquad 5^2+46=71 \qquad 19^2+96=457$. Clearly only $\boxed{(\textbf{C})}$ is our only option left. (franchester)

Solution 3

From Fermat's Little Theorom, $p^2 \equiv 1\pmod 3$ if $p$ is co-prime with $3$. So for any $n\equiv2\pmod3$, $p^2+n \equiv 0\pmod 3$ - divisible by 3, so not a prime. The only choice $\equiv2\pmod3$ is $\boxed{(\textbf{C})}$

Solution 4

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=92255"