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Difference between revisions of "2011 AMC 8 Problems/Problem 24"
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However, <math>9999</math> is clearly divisible by <math>3</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>
 
so the number of ways <math>10001</math> can be written as the sum of two primes is <math>\boxed{\textbf{(A)}\ 0}</math>
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Solution #2: The estoy regular theorem states that any number in the form <math>\text{abbba}</math> cannot be expressed as the sum of <math>2</math> primes. <math>10001</math> satisfies this condition; thus, <math>10001</math> can not be expressed as the sum of <math>2</math> primes. The answer is <math>\boxed{\textbf{(A)}\ 0}</math>.
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<math>\triangle\text{QED}</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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:37, 17 November 2014

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 $\boxed{\textbf{(A)}\ 0}$


Solution #2: The estoy regular theorem states that any number in the form $\text{abbba}$ cannot be expressed as the sum of $2$ primes. $10001$ satisfies this condition; thus, $10001$ can not be expressed as the sum of $2$ primes. The answer is $\boxed{\textbf{(A)}\ 0}$.

$\triangle\text{QED}$

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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