Find all natural two digit numbers such that when you substract by seven times the sum of its digit 
from the number you get a prime number.
1. Let the two-digit number be expressed in the form \(10a + b\), where \(a\) and \(b\) are the digits of the number, and \(a\) is the tens digit while \(b\) is the units digit.
2. We need to find all numbers such that when you subtract seven times the sum of its digits from the number, you get a prime number. Mathematically, this can be written as:
   \[
   10a + b - 7(a + b)
   \]
3. Simplify the expression:
   \[
   10a + b - 7a - 7b = 3a - 6b
   \]
4. For \(3a - 6b\) to be a prime number, it must be of the form \(3k\) where \(k\) is an integer. The only way for \(3k\) to be prime is if \(k = 1\) or \(k = -1\) (since the only prime multiples of 3 are 3 and -3).
5. Therefore, we set:
   \[
   3a - 6b = 3 \implies a - 2b = 1
   \]
6. Now, we solve for \(a\) and \(b\) such that \(a - 2b = 1\) and both \(a\) and \(b\) are digits (i.e., \(0 \leq b \leq 9\) and \(1 \leq a \leq 9\)).
7. Check each value of \(b\) from 0 to 9:
   - For \(b = 0\):
     \[
     a - 2(0) = 1 \implies a = 1
     \]
     The number is \(10 \cdot 1 + 0 = 10\).
   - For \(b = 1\):
     \[
     a - 2(1) = 1 \implies a = 3
     \]
     The number is \(10 \cdot 3 + 1 = 31\).
   - For \(b = 2\):
     \[
     a - 2(2) = 1 \implies a = 5
     \]
     The number is \(10 \cdot 5 + 2 = 52\).
   - For \(b = 3\):
     \[
     a - 2(3) = 1 \implies a = 7
     \]
     The number is \(10 \cdot 7 + 3 = 73\).
   - For \(b = 4\):
     \[
     a - 2(4) = 1 \implies a = 9
     \]
     The number is \(10 \cdot 9 + 4 = 94\).
   - For \(b = 5\) to \(b = 9\), \(a\) would be greater than 9, which is not possible since \(a\) is a single digit.

8. Therefore, the valid two-digit numbers are \(10, 31, 52, 73,\) and \(94\).

The final answer is \(\boxed{10, 31, 52, 73, 94}\).