Difference between revisions of "2011 AMC 8 Problems/Problem 24"
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The only even prime number is <math>2,</math> so our only combination could be <math>2</math> and <math>9999.</math> | The only even prime number is <math>2,</math> so our only combination could be <math>2</math> and <math>9999.</math> | ||
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 < | + | so the number of ways <math>10001</math> can be written as the sum of two primes is <math>\boxed{\textbf{(A)}\ 0}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=23|num-a=25}} | {{AMC8 box|year=2011|num-b=23|num-a=25}} | ||
Revision as of 09:15, 5 November 2012
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 
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 | ||
