Find the largest number of five different digits $\overline { a b c d e }$ such that $\overline { ab }$ , $\overline { bc }$ , $\overline { cd }$ and $\overline { de }$ are prime numbers.

Gives in response $a + b + c + d + e.$

The answer is 28.

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$\overline{ab}$ , $\overline{bc}$ , $\overline{cd}$ and $\overline{de}$ , are all prime. Since all prime numbers greater than $2$ end in $1,3,7,9$ , $(b,c,d,e) = (1,3,7,9)$ in some order. Note that $\overline{02}$ cannot be one of the prime numbers, since this would make the previous prime number end in $0$ which is impossible. Another possibility is that $\overline{ab} = \overline{02}$ , but this would produce a number with $4$ digits, and since we can find a greater example, this is not the case.

Now, since $(b,c,d,e) = (1,3,7,9)$ , we want to maximise $a$ in order to find the largest number. The largest possible value of $a$ is $8$ , and placing the other digits in descending order to maximise the number, we get the number $89731$ . By checking, this satisfies the rules, and since it is the upper bound, the answer is $8+9+7+3+1 = \boxed{28}$