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

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==Solution==
  
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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) }15}</math>.
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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
 
~IronicNinja

Revision as of 21:11, 9 January 2021

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 $n = \boxed{\textbf{(E) }27}$.

~IronicNinja

minor edit (the inclusion of not) by AlcBoy1729

Video Solution

https://youtu.be/9nWsQoYZBpk

~savannahsolver

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

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