Three-Digit Numbers Increasing on Reversal

Let the 3-digit number be $100a + 10b + c$, where $a \in \{1, 2, \dots, 9\}$, $b \in \{0, 1, \dots, 9\}$, and $c \in \{1, 2, \dots, 9\}$. Number is greater than 100.

Reversing the digits gives $100c + 10b + a$. Given: $(100c + 10b + a) - (100a + 10b + c) = 198$ $99(c - a) = 198 \implies c - a = 2$

So $c = a + 2$. Since $a \ge 1$, the possible values for $(a, c)$ are:

• $a = 1, c = 3$
• $a = 2, c = 4$
• $a = 3, c = 5$
• $a = 4, c = 6$
• $a = 5, c = 7$
• $a = 6, c = 8$
• $a = 7, c = 9$ There are $7$ possible pairs for $(a, c)$.

For each pair, $b$ can be any digit from $0$ to $9$ (10 choices). Also, check if any number $\le 100$: smallest is $103 > 100$, so all generated numbers are $> 100$.

Total such numbers $= 7 \times 10 = 70$.