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

Step-by-step solution

1. 1

   Write both numbers in expanded form. With digits $a$, $b$, $c$ (and $a\ne 0$):

   $N=100a+10b+c,\qquad N'=100c+10b+a$

2. 2

   Subtract and watch the tens digit vanish.

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

   The middle digit $b$ cancels completely, which is the whole point of the problem: it is entirely free.

3. 3

   Solve for the digit gap.

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

   So the first digit is exactly $4$ more than the last.

4. 4

   Maximise digit by digit, from the left. Place value means the hundreds digit dominates, so take the largest $a$ that keeps $c=a-4$ a valid digit: $a=9$, giving $c=5$. Then maximise the tens digit. The problem only requires the first and last digits to differ — it says nothing about the middle one — so $b$ may be $9$:

   $N=995$

5. 5

   Check, and see why 985 is not the answer.

   $995-599=396\quad\checkmark$

   A common slip is to assume all three digits must be distinct, which would rule out $b=9$ and give the smaller $985$ (also valid: $985-589=396$). But $995$ is larger and satisfies every stated condition, so it is the true maximum.