10. How many prime numbers are there such that, if any group of digits, even non-consecutive ones, is deleted from them (without deleting all of them), and the remaining digits are read in the order they are in, the result is still a prime number?

(It is reminded that 1 is not a prime number.)
(A) 7
(B) 3
(C) 5
(D) 8
(E) 10
(10) The correct answer is $(D)$.

The single-digit prime numbers are $2,3,5,7$. Therefore, any prime number with the indicated property must consist only of these digits, because removing all but one of them should result in a prime number.

Moreover, 2 and 5, if present, can only appear in the first position; otherwise, removing all digits after the 2 or 5 would result in an even number or a multiple of 5 with at least two digits, which cannot be prime. In particular, only one of the digits 2 and 5 can be present in the number, and exclusively in the first position. A digit cannot appear more than once, otherwise removing all digits except two identical ones would result in a multiple of 11, which cannot be prime.

The only two-digit numbers that satisfy the conditions described are $23,27,37,53,57,73$. Of these, only $23,37,53,73$ are prime.

For numbers with at least three digits:

- If they do not contain the digit 2 or 5, they consist only of the digits 3 and 7 and must necessarily have repeated digits. They cannot, therefore, satisfy the indicated properties.
- If they have 2 as the first digit, they cannot have 5 (which should be in the first position) or 7 (since 27 is not prime) as other digits. Each digit following the 2 is therefore a 3, but then they necessarily have repeated digits and cannot satisfy the indicated properties.
- Similarly, if they have 5 as the first digit, each subsequent digit must be a 3, and repeated digits are necessarily present.

In conclusion, the only numbers that satisfy the described property are $2,3,5,7,23,37,53,73$. Problem proposed by Emanuele Tron.