When the digits of a two-digit natural number are interchanged, the original number is greater than three times the new number obtained. How many such natural numbers satisfy this condition? Ignore numbers that have '0' in the unit place.

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**Solution:**

1. Let the original two-digit number be $10x + y$, where $x$ is the tens digit and $y$ is the units digit.

2. When digits are interchanged, the new number becomes $10y + x$.

3. Given condition:
   $$10x + y > 3(10y + x)$$

4. Expanding the right side:
   $$10x + y > 30y + 3x$$

5. Rearranging terms:
   $$10x - 3x > 30y - y$$
   $$7x > 29y$$

6. Since $x$ and $y$ are digits, $x$ ranges from 1 to 9 and $y$ ranges from 1 to 9 (units digit cannot be 0 as per the question).

7. For each $y$, find possible $x$ values satisfying:
   $$7x > 29y \ x > \frac{29y}{7}$$

8. Check for each $y$:

- For $y=1$:
     $$x > \frac{29}{7} \approx 4.14$$
     Possible $x$: 5, 6, 7, 8, 9 (5 values)

- For $y=2$:
     $$x > \frac{58}{7} \approx 8.29$$
     Possible $x$: 9 (1 value)

- For $y=3$:
     $$x > \frac{87}{7} \approx 12.43$$
     No possible $x$ since max digit is 9

- For $y \geq 4$:
     $$x > \frac{29y}{7} > 16$$
     No possible $x$

9. Total possible numbers = 5 (for $y=1$) + 1 (for $y=2$) = 6

**Answer:** (b) 6

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