MHB Sum of 2 Primes: 45 - (2 Digit Integer)?

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To solve the problem of finding the sum of two primes that results in a two-digit odd integer, the discussion focuses on subtracting the count of such integers from 45. It is established that the sum of two primes must include the prime number 2 to yield an odd result, leading to the examination of the primality of n-2 for odd integers n. The conclusion reached is that there are 21 odd two-digit numbers where n-2 is prime, resulting in a final answer of 24. The conversation highlights the importance of correctly identifying prime numbers in this context. The analysis confirms that about half of the odd two-digit numbers cannot be expressed as a sum of two primes.
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View attachment 6519 I think that we have to get all 2 digit odd numbers that can be expressed as the sum of 2 primes and subtract that from 45, so I think that the answer would be 45-(number of 2 digit integers n that are prime and have n-2 be prime as well)?
 

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Ilikebugs said:
I think that we have to get all 2 digit odd numbers that can be expressed as the sum of 2 primes and subtract that from 45, so I think that the answer would be 45-(number of 2 digit integers n that are prime and have n-2 be prime as well)?

$n$ doesn't have to be prime.
 
The question seems interesting. The answer is $24$. That is, about half of the odd 2-didit numbers cannot be expressed as a sum of two primes. The proof is simple. The sum if two primes must contain $2$, for otherwise the number cannot be odd. Thus, we have just check the primality of $n-2$ , where $n$ is any odd number.
 
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Arent there 21 odd 2 digit numbers where n-2 is prime, so its 24?
 
Ilikebugs said:
Arent there 21 odd 2 digit numbers where n-2 is prime, so its 24?

True, I missed one. Corrected though
 

Thread 'Area of Overlapping Squares'

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