A three-digit number has all three digits different. Reversing its digits gives a second number, and subtracting that from the original gives 396. Find the smallest such number.

Step-by-step solution

1. 1

   Set up the subtraction in expanded form. With digits $a$, $b$, $c$:

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

   The tens digit cancels, so the difference depends only on the first and last digits.

2. 2

   Turn the given difference into a digit condition.

   $99(a-c)=396\ \Longrightarrow\ a-c=4$

3. 3

   Minimise from the most significant digit down. To make the number as small as possible, first make $a$ as small as possible. Since $a=c+4$ and $c\ge 0$, the smallest legal choice is

   $c=0,\qquad a=4$

   Any smaller $a$ would force a negative last digit.

4. 4

   Choose the smallest tens digit that keeps all digits different. The digits $4$ and $0$ are taken, so $b$ cannot be $4$ or $0$. The next smallest available digit is

   $b=1\ \Longrightarrow\ N=410$

5. 5

   Verify the subtraction. Reversing $410$ gives the digit string $014$, i.e. the number $14$:

   $410-14=396\quad\checkmark$

   Digits $4$, $1$, $0$ are all different $\checkmark$. Compare this with the largest version of the same puzzle, where maximising instead of minimising leads to $995$ — the shared engine in both is $a-c=4$.