Difference between revisions of "2019 AMC 10B Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | 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>. | + | 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 | ~IronicNinja | ||
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==Video Solution== | ==Video Solution== | ||
| − | https://youtu.be/ | + | https://youtu.be/CgZnUftXoig |
| − | ~ | + | ~Education, the Study of Everything |
==See Also== | ==See Also== | ||
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.
Contents
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
