_ _ _ _ _ $9 \times 4 = 9 $_ _ _ _ _
I have a 6-digit positive integer whose last digit is 9. If I move
this last digit directly to the front to form another 6-digit integer,
then the original number is quadrupled.What is the original 6-digit number?
Let $\large{\overline {ABCDE9}\times4=\overline {9FGHIJ}}$
So far, using trial and error, I could figure out the list of possibilities for
$A \rightarrow2$
$B \rightarrow3,4$
$E \rightarrow3, 8,1 ,6,4,9$
$J \rightarrow6$
$I \rightarrow5,7,9$
I'm stuck at this point. I have no clue how to narrow down the set of possibilities.
EDIT
How can I possibly assume that the digits which fill the blanks of one side are identical to those of the other side?
EDIT$^{2}$
I had actually misinterpreted the question. I thought that $ABCDE$ and $FGHIJ$ are distinct.
I started reworking on the problem and came up with two different solutions, all of which are already covered by answers (of Simply Beautiful Art and Micheal Burr).
Best Answer
Hint:
$$\underbrace{x=abcde9}_{\text{we want to solve for this}}$$
$$\underbrace{900000+abcde=4x}_{\text{this is moving the 9 to the front}}$$
$$\underbrace{abcde0=x-9}_{\text{this is dropping the 9 from the end}}$$
$$\underbrace{abcde=\frac1{10}abcde0}_{\text{this is moving the digits over}}$$
Thus,
$$4x=900000+\frac1{10}(x-9)$$
Solving this gives