2011 AMC 8 Problems/Problem 24: Difference between revisions
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==Solution== | ==Solution== | ||
For the sum of two numbers to be odd, one must be odd and the other must be even. 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> so the number of ways <math>10001</math> can be written as the sum of two primes is <math>\boxed{\textbf{(A)}\ 0}</math> | For the sum of two numbers to be odd, one must be odd and the other must be even. | ||
Proof: | |||
All odd numbers are of the form <math>2n+1</math> where n is an integer. | |||
All even numbers are of the form <math>2m</math> where m is an integer. | |||
<cmath> 2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1 </cmath> and <math>m+n</math> is an integer because <math>m</math> and <math>n</math> are both integers. | |||
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> 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 11:59, 27 November 2011
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.
Proof:
All odd numbers are of the form 
where n is an integer.
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 isso 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
so our only combination could be 
and 
However, 
is clearly divisible by 
so the number of ways 
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 | ||
