Finding four digit number ABCD such that $ABCD \div DCBA = 9$

Question

If $(ABCD) \div (DCBA)=9$ where $A,B,C$ and $D$ are distinct and all them belong to ${0,1,2,3,4,5,6,7,8,9}$ but $A$ and $D $are not equal to zero then find $A,B ,C$ and $D$. I tried with the decimal expansion and I couldn't arrive at the solution

Answer 1

D has to be 1 and A has to be 9, so the problem becomes 9BC1=9x1CB9.

Either C=0 or C > 1. If C>1, 9xC would overflow to the next digit, which it can't. Therefore, C=0 and the problem becomes 9B01=9x10B9.

In values, 9001+100B = 9x1009+90B or 10B = 9081-9001=80 or B=8.

The final solution is 9801 = 9 x 1089.

Answer 2

I solved the problem in this way just for fun. Marty Cohen's method is better.

Note that the problem statement is equivalent to $ABCD=9\cdot DCBA=(10-1)DCBA=DCBA\cdot 10-DCBA=DCBA0-DCBA$

This tells us that

$$\begin{array}{cccccc} & D & C & B & A & 0 \\ - & 0 & D & C & B & A\\\hline & 0 & A & B & C & D\end{array} $$

Now note that since $D$ cannot be $0$, it must be $1$ (why?)

If you understand the hint, you should be able to continue this train of reasoning. If not,don't hesitate to ask for another hint.

Answer 3

At about midnight last night, I realized that my answer works for any base, not just decimal, so I am making this explicit here.

Let b be the base, where b >= 4. Digits in base b that are expressions are surrounded by parenthesis. The two used here are (b-1) and (b-2), which are the largest two digits in base b.

Here is the problem:

If ABCD and DCBA are 4-digit numbers in base B with all digits distinct and A and D non-zero, and ABCD = (b-1) x DCBA ("x" indicates multiplication), determine the numbers.

Solution:

If D > 1, the product would have five digits. Therefore D=1 and A=(b-1).

If C>1, the product (b-1)xC would overflow and, again, the product would be five digits. Since 1 is already taken, C=0.

The problem is now (b-1) B 0 1 = (b-1) x 1 0 B (b-1). Getting rid of the high-order digit, this becomes B 0 1 = (b-1) x B (b-1). Converting to values, Bb^2+1 = (b-1)(Bb+b-1) =Bb^2-Bb+b^2-2b+1 or Bb = b^2-2b =b(b-2) or B = (b-2).

The solution is therefore

(b-1)(b-2) 0 1 = (b-1) x 1 0 (b-2) (b-1).