Difference between revisions of "2019 AMC 10B Problems/Problem 2"

(Undo revision 102900 by Xmidnightfirex (talk))
(Tag: Undo)
m (Fixed grammar and removed multiple irrelevant attributions)
Line 9: Line 9:
 
==Solution==
 
==Solution==
  
Since a counterexample must be when <math>n</math> is not prime, <math>n</math> must be composite, so we eliminate A and C. Now we subtract <math>2</math> from the remaining answer choices, and we see that the only time <math>n-2</math> is '''not''' prime is when <math>n = 27</math>, which is <math>\boxed{\textbf{(E) }27}</math>.
+
Since a counterexample must be value of <math>n</math> which is not prime, <math>n</math> must be composite, so we eliminate <math>\text{A}</math> and <math>\text{C}</math>. Now we subtract <math>2</math> from the remaining answer choices, and we see that the only time <math>n-2</math> is '''not''' prime is when <math>n = \boxed{\textbf{(E) }27}</math>.
 
 
~IronicNinja
 
 
 
minor edit (the inclusion of not) by AlcBoy1729
 
  
 
==See Also==
 
==See Also==

Revision as of 19:30, 18 February 2019

The following problem is from both the 2019 AMC 10B #2 and 2019 AMC 12B #2, so both problems redirect to this page.

Problem

Consider the statement, "If $n$ is not prime, then $n-2$ is prime." Which of the following values of $n$ is a counterexample to this statement?

$\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27$

Solution

Since a counterexample must be value of $n$ which is not prime, $n$ must be composite, so we eliminate $\text{A}$ and $\text{C}$. Now we subtract $2$ from the remaining answer choices, and we see that the only time $n-2$ is not prime is when $n = \boxed{\textbf{(E) }27}$.

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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
2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 12 Problems and Solutions

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