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

Revision as of 23:30, 19 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 (Not Recommended Solution)

We proceed with guess and check: $3^2+16=25 \qquad 1^2+24=25 \qquad 1^2+46=71 \qquad 5^2+96=121$ . Clearly only $\boxed{(\textbf{C})}$ is our only option left. -liu4505

Solution 3

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 $\boxed{\text{(C)}}$ .

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 13: / Line 13: @@
 ^2+24=25 \qquad
 ^2+46=71 \qquad
-^2+96=457</math>.
+^2+96=121</math>.
 Clearly only <math>\boxed{(\textbf{C})}</math> is our only option left.
 -liu4505

Page

Toolbox

Search

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

Revision as of 23:30, 19 January 2019

Contents

Solution 1

Solution 2 (Not Recommended Solution)

Solution 3

See Also