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

(Solution)
(Solution)
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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>
 
 
 
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>.
 
 
<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 21:12, 18 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 $\boxed{\textbf{(A)}\ 0}$

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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