A three-digit number has all distinct digits. Subtracting the number written with the same digits in reverse order gives 693. Find the smallest such number.

Step-by-step solution

1. 1

   Write both numbers in place-value form. Let the digits be $a$ (hundreds), $b$ (tens), $c$ (units), so the number is $100a + 10b + c$ and its reverse is $100c + 10b + a$.

2. 2

   Subtract and watch the tens digit vanish.

   $(100a + 10b + c) - (100c + 10b + a) = 99a - 99c = 99(a-c)$

   The middle digit cancels entirely — every such difference is a multiple of $99$, and $b$ never influences it.

3. 3

   Solve for the digit gap.

   $99(a-c) = 693 \quad\Longrightarrow\quad a - c = \frac{693}{99} = 7$

4. 4

   List the digit pairs and pick the smallest hundreds digit. With $a, c \in \{0,\dots,9\}$ and $a - c = 7$, the options are $(a,c) = (7,0), (8,1), (9,2)$. To minimise the number, the hundreds digit must be as small as possible, so take $a = 7$, $c = 0$.

5. 5

   Choose the smallest admissible tens digit. The number is $7b0$, and all three digits must differ, so $b \ne 7$ and $b \ne 0$. The smallest remaining choice is $b = 1$, giving

   $710$

6. 6

   Verify. The reverse of $710$ is $017 = 17$, and $710 - 17 = 693$. The digits $7, 1, 0$ are distinct. Any smaller candidate would need $a \le 6$, which makes $a - c = 7$ impossible since $c \ge 0$; and $700$ has repeated digits while $710$ is the next number of the form $7b0$.