How many unordered pairs of prime numbers have a sum of 40?
To find the unordered pairs of prime numbers that have a sum of 40, I need to look for prime numbers that are less than 40 and see if their complement is also a prime number. The smallest prime number is 2, and its complement is 40 - 2 = 38, which is not prime. The next prime number is 3, and its complement is 40 - 3 = 37, which is prime. So {3, 37} is one unordered pair. The next prime number is 5, and its complement is 40 - 5 = 35, which is not prime. The next prime number is 7, and its complement is 40 - 7 = 33, which is not prime. The next prime number is 11, and its complement is 40 - 11 = 29, which is prime. So {11, 29} is another unordered pair. The next prime number is 13, and its complement is 40 - 13 = 27, which is not prime. The next prime number is 17, and its complement is 40 - 17 = 23, which is prime. So {17, 23} is another unordered pair. The next prime number is 19, and its complement is 40 - 19 = 21, which is not prime. The next prime number is 31, and its complement is 40 - 31 = 9, which is not prime. The next prime number is 41, which is larger than 40, so I don't need to consider it. Therefore, the only unordered pairs of prime numbers that have a sum of 40 are {3, 37}, {11, 29}, and {17, 23}. There are 3 such pairs. 