Difference between revisions of "2011 AMC 8 Problems/Problem 24"
Huangjialan (talk | contribs) (→Solution 3(Simple)) |
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.) |
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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(Simple)== | |
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− | ==Solution | ||
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 01:52, 21 October 2023
Contents
[hide]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. 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(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 |
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