Find all the values that can take the last digit of a "perfect" even number. (The natural number $n$ is called "perfect" if the sum of all its natural divisors is equal twice the number itself.For example: the number $6$ is perfect ,because $1+2+3+6=2\cdot6$).
1. **Definition and Initial Observations**:
   - A natural number \( n \) is called "perfect" if the sum of all its natural divisors is equal to twice the number itself. Mathematically, if \( \sigma(n) \) denotes the sum of the natural divisors of \( n \), then \( n \) is perfect if \( \sigma(n) = 2n \).
   - The first couple of perfect numbers are 6 and 28, both of which are even.

2. **Form of Even Perfect Numbers**:
   - Assume any general even perfect number can be written in the form \( n = 2^k p \) where \( p \) is a prime number.
   - Given \( \sigma(n) = 2n \), we have \( \sigma(2^k p) = 2^{k+1} p \).

3. **Properties of Divisors**:
   - Note that \( \gcd(2^k, p) = 1 \), so \( \sigma(2^k p) = \sigma(2^k) \sigma(p) \).
   - The sum of divisors of \( 2^k \) is \( \sigma(2^k) = 1 + 2 + 2^2 + \dots + 2^k = 2^{k+1} - 1 \).
   - The sum of divisors of \( p \) is \( \sigma(p) = p + 1 \).

4. **Equating the Sum of Divisors**:
   - We obtain \( \sigma(2^k p) = (2^{k+1} - 1)(p + 1) \).
   - Given \( \sigma(2^k p) = 2^{k+1} p \), we equate:
     \[
     (2^{k+1} - 1)(p + 1) = 2^{k+1} p
     \]

5. **Solving for \( p \)**:
   - Dividing both sides by \( p + 1 \):
     \[
     \frac{p}{p + 1} = \frac{2^{k+1} - 1}{2^{k+1}}
     \]
   - Note that on both the left and right-hand sides of the equation, the numerator is one less than the denominator. Thus, a solution for \( p \) would be:
     \[
     p = 2^{k+1} - 1
     \]

6. **Form of \( n \)**:
   - Then \( n = 2^k (2^{k+1} - 1) \) where \( 2^{k+1} - 1 \) is prime (called a Mersenne prime).
   - We can shift the \( k \)'s by one to obtain:
     \[
     n = 2^{k-1} (2^k - 1)
     \]
   - Here, \( 2^k - 1 \) must be prime, and this only occurs for certain values of \( k \) which are prime.

7. **Last Digit of Even Perfect Numbers**:
   - To find the last digit of \( n = 2^{k-1} (2^k - 1) \), we need to consider the last digits of powers of 2 and Mersenne primes.
   - The last digits of powers of 2 cycle every 4: \( 2, 4, 8, 6 \).
   - The last digit of \( 2^k - 1 \) depends on \( k \mod 4 \):
     - If \( k \equiv 1 \mod 4 \), \( 2^k \equiv 2 \mod 10 \), so \( 2^k - 1 \equiv 1 \mod 10 \).
     - If \( k \equiv 2 \mod 4 \), \( 2^k \equiv 4 \mod 10 \), so \( 2^k - 1 \equiv 3 \mod 10 \).
     - If \( k \equiv 3 \mod 4 \), \( 2^k \equiv 8 \mod 10 \), so \( 2^k - 1 \equiv 7 \mod 10 \).
     - If \( k \equiv 0 \mod 4 \), \( 2^k \equiv 6 \mod 10 \), so \( 2^k - 1 \equiv 5 \mod 10 \).

8. **Combining the Last Digits**:
   - The last digit of \( n = 2^{k-1} (2^k - 1) \) will be the product of the last digits of \( 2^{k-1} \) and \( 2^k - 1 \).
   - Considering the cycles and possible last digits, the possible last digits of even perfect numbers are 6 and 8.

The final answer is \(\boxed{6 \text{ and } 8}\).