Difference between revisions of "2011 AMC 8 Problems/Problem 24"

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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, 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.
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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.
 
<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>  
 
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>  
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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 <cmath>\boxed{\textbf{(A)}\ 0}</cmath>
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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}}
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{{MAA Notice}}

Revision as of 00:53, 7 November 2020

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

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