2019 AMC 10B Problems/Problem 2: Difference between revisions
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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #2]] and [[2019 AMC 12B Problems|2019 AMC 12B #2]]}} | |||
==Problem== | |||
Consider the statement, "If <math>n</math> is not prime, then <math>n-2</math> is prime." Which of the following values of <math>n</math> is a counterexample to this statement? | |||
<math>\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27</math> | |||
==Solution== | |||
Since a counterexample must be a 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 | |||
==Video Solution== | |||
https://youtu.be/CgZnUftXoig | |||
~Education, the Study of Everything | |||
==See Also== | |||
{{AMC10 box|year=2019|ab=B|num-b=1|num-a=3}} | |||
{{AMC12 box|year=2019|ab=B|num-b=1|num-a=3}} | |||
{{MAA Notice}} | |||
Latest revision as of 09:23, 24 June 2023
- 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 
is not prime, then 
is prime." Which of the following values of is a counterexample to this statement?
Solution
Since a counterexample must be a value of which is not prime, must be composite, so we eliminate 
and 
. Now we subtract 
from the remaining answer choices, and we see that the only time is not prime is when 
.
~IronicNinja
minor edit (the inclusion of not) by AlcBoy1729
Video Solution
~Education, the Study of Everything
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
