Prepostorous Primes

Note that two-digit primes can only end with $1,3,7,9$ . Since all $a,b,c$ are last digits (with $\overline{ab}, \overline{bc}, \overline{ca}$ for example), they must all be $1,3,7,9$ .

The digits cannot be all among $1$ and $7$ , as then $\overline{abc} = 100a + 10b + c \equiv a+b+c \equiv 1+1+1 \equiv 0 \pmod{3}$ , and so $\overline{abc}$ is divisible by $3$ , impossible. Also, we cannot have two digits divisible by $3$ . Suppose otherwise, that $a,b$ are divisible by $3$ , then $\overline{ab}$ is divisible by $3$ by a similar argument as above. So we need at least one digit divisible by $3$ (otherwise all digits are among $1,7$ , impossible), but no more than one digit (otherwise we have at least two digits divisible by $3$ , impossible).

Now, ignore the condition that $\overline{abc}$ is prime; after we get all possible candidates for $\overline{abc}$ , only then we check whether each is prime.

Suppose that some of the digits is $9$ ; WLOG $a = 9$ . The only two-digit prime beginning with $9$ is $97$ . Since $\overline{ab}, \overline{ac}$ are both primes, we need $b = c = 7$ . But then $\overline{bc} = 77$ is not prime. So none of the digits is $9$ . Thus the digit divisible by $3$ must be a $3$ .

Among the two remaining digits, we cannot have both $7$ s, as that means we have a $77$ which is not prime. However, we can check that $113$ and $173$ , together with permutations, satisfy the conditions that all 6 two-digit primes among them are primes. Thus we only need to check whether nine numbers are prime. Of course, this is a job for computer, because I don't feel like doing this boring task.

Thus the sum is $113 + 131 + 311 + 137 + 173 + 317 = \boxed{1182}$ .

My solution is logic and table of prime number 2, 4, 5, 6 and 8 are exactly impossible!

so, only 1, 3, 7, 9

WE know that 9 is also impossible because, from all the other numbers that combine with 9, if we just take 2 of the digits abc, it must be not prime. Whether I take 9 and 7, both of them when bonds are prime, any numbers will make them not prime when we bond 2 of three numbers. so, the possible answer is digit 1, 3 and 7

The numbers fulfil (after I filter) are 113 131 137 173 311 317

Add them, got 1182