Three palaces, each rotating on a duck leg, make a full round in $30$, $50$, and $70$ days, respectively. Today, at noon, all three palaces face northwards. In how many days will they all face southwards?
1. **Understanding the Problem:**
   - Each palace makes a full rotation in 30, 50, and 70 days respectively.
   - Today, all palaces face northwards.
   - We need to find the smallest \( x \) such that all palaces face southwards.

2. **Determine the Southward Facing Days:**
   - For the first palace (30 days cycle), it faces southwards after 15 days.
   - For the second palace (50 days cycle), it faces southwards after 25 days.
   - For the third palace (70 days cycle), it faces southwards after 35 days.

3. **Set Up the Congruences:**
   - \( x \equiv 15 \pmod{30} \)
   - \( x \equiv 25 \pmod{50} \)
   - \( x \equiv 35 \pmod{70} \)

4. **Solve the System of Congruences:**
   - We use the Chinese Remainder Theorem (CRT) to solve this system.

5. **Chinese Remainder Theorem Application:**
   - The moduli 30, 50, and 70 are not pairwise coprime, so we need to find a common modulus.
   - The least common multiple (LCM) of 30, 50, and 70 is 1050.
   - We need to find \( x \) such that:
     \[
     x \equiv 15 \pmod{30}
     \]
     \[
     x \equiv 25 \pmod{50}
     \]
     \[
     x \equiv 35 \pmod{70}
     \]

6. **Simplify the Congruences:**
   - Since \( x \equiv 15 \pmod{30} \), we can write \( x = 30k + 15 \) for some integer \( k \).
   - Substitute \( x = 30k + 15 \) into the second congruence:
     \[
     30k + 15 \equiv 25 \pmod{50}
     \]
     \[
     30k \equiv 10 \pmod{50}
     \]
     \[
     3k \equiv 1 \pmod{5}
     \]
   - The multiplicative inverse of 3 modulo 5 is 2 (since \( 3 \cdot 2 \equiv 1 \pmod{5} \)).
   - Thus, \( k \equiv 2 \pmod{5} \), so \( k = 5m + 2 \) for some integer \( m \).
   - Substitute \( k = 5m + 2 \) back into \( x = 30k + 15 \):
     \[
     x = 30(5m + 2) + 15
     \]
     \[
     x = 150m + 60 + 15
     \]
     \[
     x = 150m + 75
     \]

7. **Substitute into the Third Congruence:**
   - Substitute \( x = 150m + 75 \) into the third congruence:
     \[
     150m + 75 \equiv 35 \pmod{70}
     \]
     \[
     150m \equiv -40 \pmod{70}
     \]
     \[
     150m \equiv 30 \pmod{70}
     \]
     \[
     15m \equiv 3 \pmod{7}
     \]
   - The multiplicative inverse of 15 modulo 7 is 1 (since \( 15 \equiv 1 \pmod{7} \)).
   - Thus, \( m \equiv 3 \pmod{7} \), so \( m = 7n + 3 \) for some integer \( n \).
   - Substitute \( m = 7n + 3 \) back into \( x = 150m + 75 \):
     \[
     x = 150(7n + 3) + 75
     \]
     \[
     x = 1050n + 450 + 75
     \]
     \[
     x = 1050n + 525
     \]

8. **Conclusion:**
   - The smallest positive \( x \) is when \( n = 0 \):
     \[
     x = 525
     \]

The final answer is \( \boxed{525} \)