Difference between revisions of "2011 AMC 8 Problems/Problem 24"
(Problem 24) |
|||
| Line 1: | Line 1: | ||
| − | In how many ways can 10001 be written as the sum of two primes? | + | In how many ways can <math>10001</math> be written as the sum of two primes? |
<math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4 </math> | <math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4 </math> | ||
==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> | ||
==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 20:04, 25 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. 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 | ||
