Difference between revisions of "2011 AMC 8 Problems/Problem 24"
Scrabbler94 (talk | contribs) (Solution 2 is completely wrong. If this were to be applied to the number 100001 (31 if interpreted in binary), 31 can be written as the sum of 2 primes as 2+29, but 99999 is not prime.) |
Scrabbler94 (talk | contribs) (Undo revision 199921 by Scrabbler94 (talk)) (Tag: Undo) |
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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> | ||
| + | ==Solution 2 (Sort of)== | ||
| + | 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>. | ||
| − | ==Solution | + | 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. |
| + | |||
| + | ==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 01:54, 21 October 2023
Contents
Problem
In how many ways can 
be written as the sum of two primes?
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 
where n is an integer, and all even numbers are of the form 
where m is an integer.
![\[2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1\]](http://latex.artofproblemsolving.com/9/3/3/933cf5ae735e7ffc000e78c87f9efcf317d38ab8.png)
and 
is an integer because 
and 
are both integers.
The only even prime number is 
so our only combination could be 
and 
However, 
is clearly divisible by 
,
so the number of ways can be written as the sum of two primes is 
Solution 2 (Sort of)
One interesting way to do this is to think of as if it's binary. Converting it to base 
would result in the number 
. Since cannot be written as the sum of two primes, the answer is 
.
Note: This is not a valid way to do problems like this. For example, the number 
can be written as the sum of two primes in 
ways, but if we convert to base ten, we would get 
which obviously cannot be written as the sum of two primes in ways.
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
Video Solution
https://youtu.be/qJuoLucUn9o by David
Video Solution 2
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
