Sum Of 2 Primes

Since all of the options have odd numbers as their numbers, and only even $+$ odd = odd, the only prime number which is even is 2. Now we can try checking each of the options. $41 - 2 = 39$ , which is incorrect because $39 = 3 \times 13$ , so $39$ is not a prime number. $39 - 2 = 37$ , which is correct because $37$ is a prime number. So the answer is $\boxed{39}$ .

Short solution

Notice that sum of any two prime numbers is always an even number . Since the sum of two primes is odd . Therefore, one of the prime must be even which is clearly 2. Then $S= p+2 \implies p = S-2$ From the given sums we find $p= 37\Rightarrow S = \boxed{39}$ .

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Explained solution

For any prime number $\,(p>5)$ can be expressed as $p = 6k \pm 1\phantom{cc} k\in\mathbb N$ Say two primes $\,(p_1 ,p_2)$ are greater than 5. Then following above formula (not a general formula )we can have $S = p_1+p_2 = \,(6k_1\pm 1 )+\,(6k_2 \pm 1)= 2\left\{3\,(k_1+k_2)\pm 1\right\}\phantom{c} \text{or} \phantom{c}6(k_1+k_2)$ Shows that sum of two primes is an even number and among given $S$ , we have only odd numbers. From here can conclude that one them prime $\,(p_1 <5)$ and $p_2>5$ since maximum $S=41$ .

Obviously, $p_1=2$ and then $S = 2 + 6k_2\pm 1 \implies k_2 = \dfrac{S-2\pm 1 }{6}= \begin{cases} \dfrac{S-1}{6} = \dfrac{34}{6}, \dfrac{38}{6} , \dfrac{40}{6} , 6\\\dfrac{S-3}{6}=\dfrac{32}{6}, 6 , \dfrac{38}{6} , \dfrac{34}{6}\end{cases}$ We obtained $k_2 =6 \implies p_2 = 6\times 6\pm 1 = 35 , 37$ and possible sum we can obtained is $S = 37+2=39$ .