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

(Undo revision 199921 by Scrabbler94 (talk))
(Tag: Undo)
(Solution 2 is completely wrong. If this were to be applied to the number 100001 (33 if interpreted in binary), 33 can be written as the sum of 2 primes as 2+31, but 99999 is not prime.))
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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>
 
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 (Sort of)==
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==Solution 2(Simple)==
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One interesting way to do this is to think of <math>10001</math> as if it's binary. Converting it to base <math>10</math> would result in the number <math>17</math>. Since <math>17</math> cannot be written as the sum of two primes, the answer is <math>\boxed{\textbf{(A)} 0}</math>.
 
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Note: This is not a valid way to do problems like this. For example, the number <math>1000</math> can be written as the sum of two primes in <math>28</math> ways, but if we convert <math>1000</math> to base ten, we would get <math>16</math> which obviously cannot be written as the sum of two primes in <math>28</math> ways.
 
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==Solution 3(Simple)==
 
 
First, we noticed that 10001 is equal to 5000+5001, if you subtract n to 5000 and add n to 5001, you always get an even number, even number is never a prime number except 2. We try 2 and 9999 but we can see 9999 is divisible by 3 and 9 clearly. So the answer is <math>\boxed{\textbf{(A)} 0}</math>
 
First, we noticed that 10001 is equal to 5000+5001, if you subtract n to 5000 and add n to 5001, you always get an even number, even number is never a prime number except 2. We try 2 and 9999 but we can see 9999 is divisible by 3 and 9 clearly. So the answer is <math>\boxed{\textbf{(A)} 0}</math>
  

Revision as of 01:55, 21 October 2023

Problem

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(Simple)

First, we noticed that 10001 is equal to 5000+5001, if you subtract n to 5000 and add n to 5001, you always get an even number, even number is never a prime number except 2. We try 2 and 9999 but we can see 9999 is divisible by 3 and 9 clearly. So the answer is $\boxed{\textbf{(A)} 0}$

Video Solution

https://youtu.be/qJuoLucUn9o by David

Video Solution 2

https://youtu.be/GqTHx0tOB4o

~savannahsolver

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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