# Multiplication

• November 22nd 2010, 04:17 PM
matgrl
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?
• November 22nd 2010, 04:51 PM
wonderboy1953
Quote:

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?
• November 29th 2010, 06:48 PM
matgrl
yes please and thank you i do not know how to prove this
• November 29th 2010, 06:55 PM
Ackbeet
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?
• November 29th 2010, 09:27 PM
Soroban
Hello, matgrl!

Quote:

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

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

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

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

So we have:

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

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

Then we have:

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

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

Therefore, the solution is:

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

• December 1st 2010, 10:08 AM
matgrl
wonderful thank you!