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

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==Solution==
 
==Solution==
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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.
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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>  

Revision as of 23:53, 6 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 (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

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