Difference between revisions of "2011 AMC 8 Problems/Problem 24"
| Line 11: | Line 11: | ||
However, <math>9999</math> is clearly divisible by <math>3</math>, | However, <math>9999</math> is clearly divisible by <math>3</math>, | ||
so the number of ways <math>10001</math> can be written as the sum of two primes is <math>\boxed{\textbf{(A)}\ 0}</math> | so the number of ways <math>10001</math> can be written as the sum of two primes is <math>\boxed{\textbf{(A)}\ 0}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/qJuoLucUn9o | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=23|num-a=25}} | {{AMC8 box|year=2011|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:46, 7 October 2021
Contents
Problem
In how many ways can 
be written as the sum of two primes?
Solution
For the sum of two numbers to be odd, one must be odd and the other must be even, because all odd numbers are of the form 
where n is an integer, and all even numbers are of the form 
where m is an integer.
![\[2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1\]](http://latex.artofproblemsolving.com/9/3/3/933cf5ae735e7ffc000e78c87f9efcf317d38ab8.png)
and 
is an integer because 
and 
are both integers.
The only even prime number is 
so our only combination could be 
and 
However, 
is clearly divisible by 
,
so the number of ways can be written as the sum of two primes is 
Video Solution
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
