Difference between revisions of "2016 AMC 12A Problems/Problem 5"
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In this case, a counterexample is a number that would prove Goldbach's conjecture false. The conjecture asserts what '''can''' be done with '''even''' integers greater than 2. | In this case, a counterexample is a number that would prove Goldbach's conjecture false. The conjecture asserts what '''can''' be done with '''even''' integers greater than 2. | ||
Therefore the solution is <cmath>{\textbf{(E)}\text{ an even integer greater than 2 that cannot be written as the sum of two prime numbers.}}</cmath> | Therefore the solution is <cmath>{\textbf{(E)}\text{ an even integer greater than 2 that cannot be written as the sum of two prime numbers.}}</cmath> | ||
| + | |||
| + | ==Note== | ||
| + | Goldbach's conjecture has been proven for all integers less than <math>4 \cdot 10^{18}</math>, but the conjecture remains open. For more information, see [https://en.wikipedia.org/wiki/Goldbach%27s_conjecture the Wikipedia article]. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=A|num-b=4|num-a=6}} | {{AMC12 box|year=2016|ab=A|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 14:58, 11 October 2025
Contents
Problem
Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, 
). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?
Solution
In this case, a counterexample is a number that would prove Goldbach's conjecture false. The conjecture asserts what can be done with even integers greater than 2.
Therefore the solution is ![\[{\textbf{(E)}\text{ an even integer greater than 2 that cannot be written as the sum of two prime numbers.}}\]](http://latex.artofproblemsolving.com/d/0/1/d01c4e58ff9cbe484d3837f31384574d5e6fa60f.png)
Note
Goldbach's conjecture has been proven for all integers less than 
, but the conjecture remains open. For more information, see the Wikipedia article.
See Also
| 2016 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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. 
