2011 AMC 8 Problems/Problem 24
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. 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 also see that whenever an addend is an odd number, the other addend will be even, so having an odd number as an addend is not possible, other than 9999 and 2, because 2 is a prime. We try 2 and 9999 but we can see 9999 is divisible by 3 and 9 clearly. So the answer is
Solution 3 (assumed previous knowledge)
It is helpful to know and understand the Goldbach Conjecture - that every even number can be written as the sum of primes - and also, that the that are the sum of two primes are exactly two more than a prime. This is because to make the sum of two numbers odd, you must have one even and one odd. There is only one even prime, which is two, so the sum will be of the form . Hence, the odd numbers that are the sum of two primes are exactly more than a prime. Relating to the problem, is not more than a prime, because and we can easily see that is divisible by . Therefore, cannot be written as the sum of two primes, and the answer is ~mk
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.