1. ## Multiplication

Multiplication by 9 reverses a four-digit number (produces the four-digit number with the same digits in the reverse order.) What is the number?

I know the answer is 1089*9 = 9801 but there is a method to this madness. Does anyone know how to solve this?

2. Originally Posted by matgrl
Multiplication by 9 reverses a four-digit number (produces the four-digit number with the same digits in the reverse order.) What is the number?

I know the answer is 1089*9 = 9801 but there is a method to this madness. Does anyone know how to solve this?
Solve what? You just did it. Do you mean to prove there's no other possible answer?

3. yes please and thank you i do not know how to prove this

4. Why not try something like setting up the following equations:

9(a + 10b + 100c + 1000d) = d + 10c + 100b + 1000a (digit reversing)
a+b+c+d = 9n, for some integer n (divisibility by 9). Either n = 1, 2, 3, or 4, because of the following constraints:

0 <= a,b,c,d <= 9.

Maybe that could get you started?

5. Hello, matgrl!

$\displaystyle \text{Multiplication by 9 reverses a four-digit number (produces}$
$\displaystyle \text{a four-digit number with the same digits in reverse order.)}$
$\displaystyle \text{What is the number?}$

I will assume that $\displaystyle A,B,C,D$ are four different digits.

. . $\displaystyle \begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\ A & B & C & D \\ \times &&& 9 \\ \hline D & C & B & A \end{array}$

In column-1, we see that: .$\displaystyle A = 1,\:D = 9$
And there is no "carry" from column-2.

So we have:

. . $\displaystyle \begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\ 1 & B & C & 9 \\ \times &&& 9 \\ \hline 9 & C & B & 1 \end{array}$

Since there is no "carry" frim column-2, $\displaystyle \,B$ must be 0 or 1.
Since $\displaystyle A = 1$, then: .$\displaystyle B \,=\,0$

Then we have:

. . $\displaystyle \begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\ 1 & 0 & C & 9 \\ \times &&& 9 \\ \hline 9 & C & 0 & 1 \end{array}$

In column-3, we have: .$\displaystyle 9C + 8$ ends in $\displaystyle 0.$
. . Hence, $\displaystyle \,9C$ ends in $\displaystyle 2 \quad\Rightarrow\quad C = 8$

Therefore, the solution is:

. . $\displaystyle \begin{array}{cccc} 1 & 0 & 8 & 9 \\ \times &&& 9 \\ \hline 9 & 8 & 0 & 1 \end{array}$

6. wonderful thank you!