Petya gave Vasya a number puzzle. Petya chose a digit $X$ and said, "I am thinking of a three digit number that is divisible by 11. The hundreds digit is $X$ and the tens digit is 3. Find the units digit." Vasya was excited because he knew how to solve this problem, but then realized that the problem Petya gave did not have an answer. What digit $X$ did Petya chose?

[i]Ray Li.[/i]
1. Let \( K = X3Y \) be the three-digit number where \( X \) is the hundreds digit, 3 is the tens digit, and \( Y \) is the units digit.
2. A number is divisible by 11 if and only if the alternating sum of its digits is divisible by 11. For the number \( K = X3Y \), the alternating sum is given by:
   \[
   X - 3 + Y
   \]
3. For \( K \) to be divisible by 11, the alternating sum must satisfy:
   \[
   X - 3 + Y \equiv 0 \pmod{11}
   \]
   This can be rewritten as:
   \[
   X + Y - 3 \equiv 0 \pmod{11}
   \]
   or equivalently:
   \[
   X + Y \equiv 3 \pmod{11}
   \]
4. We need to find a digit \( X \) such that there is no possible digit \( Y \) (where \( 0 \leq Y \leq 9 \)) that satisfies the equation \( X + Y \equiv 3 \pmod{11} \).

5. Let's test each possible value of \( X \) from 0 to 9:
   - If \( X = 0 \), then \( Y \equiv 3 \pmod{11} \) which gives \( Y = 3 \) (valid).
   - If \( X = 1 \), then \( Y \equiv 2 \pmod{11} \) which gives \( Y = 2 \) (valid).
   - If \( X = 2 \), then \( Y \equiv 1 \pmod{11} \) which gives \( Y = 1 \) (valid).
   - If \( X = 3 \), then \( Y \equiv 0 \pmod{11} \) which gives \( Y = 0 \) (valid).
   - If \( X = 4 \), then \( Y \equiv -1 \pmod{11} \) which gives \( Y = 10 \) (not a valid digit).
   - If \( X = 5 \), then \( Y \equiv -2 \pmod{11} \) which gives \( Y = 9 \) (valid).
   - If \( X = 6 \), then \( Y \equiv -3 \pmod{11} \) which gives \( Y = 8 \) (valid).
   - If \( X = 7 \), then \( Y \equiv -4 \pmod{11} \) which gives \( Y = 7 \) (valid).
   - If \( X = 8 \), then \( Y \equiv -5 \pmod{11} \) which gives \( Y = 6 \) (valid).
   - If \( X = 9 \), then \( Y \equiv -6 \pmod{11} \) which gives \( Y = 5 \) (valid).

6. From the above, we see that the only value of \( X \) for which there is no valid digit \( Y \) that satisfies the equation \( X + Y \equiv 3 \pmod{11} \) is \( X = 4 \).

\(\blacksquare\)

The final answer is \( \boxed{ X = 4 } \)