What five digit number when multiplied by 4 produces a five digit number whose digits are those of the original number but in reverse order?
The leading digit can only be 1 or 2, and as when multiplied by 4 you get the
digits reversed it cannout be 1 (as then we would have an even number ending in 1),
so the number is of the form 2 . . . 8
Now the next most significant digit must be 0, 1 or 2 (as there is no carry when
multiplied by 4), and trial and error show that 21978 * 4 = 87912
RonL
Hello, Ian!
Here is a start on the problem . . .
The problem looks like this:What five-digit number, when multiplied by 4, produces a five digit-number
whose digits are those of the original number but in reverse order?
. . . $\displaystyle \begin{array}{ccccc}^1 & ^2 & ^3 & ^4 & ^5 \\
A & B & C & D & E \\
\times & & & & 4 \\ \hline
E & D & C & B & A \end{array}$
In column-1: .$\displaystyle 4\times A \:\to \:E$ . . . Hence: .$\displaystyle A \:=\:1\text{ or }2$
In column-5: .$\displaystyle 4 \times E$ is even . . . Hence: .$\displaystyle A \:=\:2$
. . . $\displaystyle \begin{array}{ccccc}^1 & ^2 & ^3 & ^4 & ^5 \\
2 & B & C & D & E \\
\times & & & & 4 \\ \hline
E & D & C & B & 2 \end{array}$
In column-1: .$\displaystyle E \:=\:8\text{ or }9$
In column-1: .$\displaystyle 4 \times E \:\to\:2$ . . . Hence: .$\displaystyle E \:= \:8$
. . . $\displaystyle \begin{array}{ccccc}^1 & ^2 & ^3 & ^4 & ^5 \\
2 & B & C & D & 8 \\
\times & & & & 4 \\ \hline
8 & D & C & B & A \end{array}$
In column-2: .$\displaystyle 4 \times B \to D$ ... and there is no "carry" to column-1.
. . Hence: .$\displaystyle B \:=\:0\text{ or }1$
Can you finish it?