Count of two-digit numbers

Let the two-digit number be $10a + b$, where $a, b \in \{1, 2, \dots, 9\}$. Interchanging the digits gives $10b + a$.

Given $10a + b > 3(10b + a)$: $10a + b > 30b + 3a$ $7a > 29b \implies a > \frac{29}{7}b$

Since $a \le 9$ and $b \ge 1$:

• If $b = 1$: $a > \frac{29}{7} \approx 4.14 \implies a \in \{5, 6, 7, 8, 9\}$ (5 numbers).
• If $b = 2$: $a > \frac{58}{7} \approx 8.28 \implies a = 9$ (1 number).
• If $b \ge 3$: $\frac{29}{7}(3) > 12 > 9$, no possible digits for $a$.

Total numbers $= 5 + 1 = 6$.