Art of Problem Solving

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.
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 <math>2n+1</math> where n is an integer, and all even numbers are of the form <math>2m</math> where m is an integer.
 
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.
<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>
  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 $10001$ be written as the sum of two primes?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

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 $2n+1$ where n is an integer, and all even numbers are of the form $2m$ where m is an integer. \[2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1\] and $m+n$ is an integer because $m$ and $n$ are both integers.

The only even prime number is $2,$ so our only combination could be $2$ and $9999.$ However, $9999$ is clearly divisible by $3$ so the number of ways $10001$ can be written as the sum of two primes is $\boxed{\textbf{(A)}\ 0}$

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
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