Moving the last digit of a number to the front

Easy solution: let x be the 5-digit number such that x9 (adding a 9 to it) gives the required number. Then $x$ satisfies
$900000 + x = 4 \times (10x + 9) \implies x = 23076$

Just an alternative solution. (Personally, I prefer the elegance of the solution by @H K )

Multiply 4 by the unit-digit in the original number ( $\color{#D61F06}9$ ).

$9\times4=36$

The last digit of 36 is 6.

Which becomes the 10-digit in the original number ( $\color{#D61F06}6$ 9).

$69\times4=276$

The 2nd last digit of 276 is 7.

Which becomes the 100-digit in the original number ( $\color{#D61F06}7$ 69).

$769\times4=3076$

The 3rd last digit of 3076 is 0. When a 0 is the result, the preceding digit is also used (30).

Which becomes the 10,000-digit and 1,000-digit in the original number ( $\color{#D61F06}30$ 769).

$30769\times4=123076$

The 5th last digit of 123076 is 2.

Which becomes the 100,000-digit in the original number ( $\color{#D61F06}2$ 30769).

$\large \boxed{230769} \times4=923076$

[EDIT] For those having difficulty with the phrasing of this problem:

'Move' is not a mathematical function, it's a physical operation, so we should conceptualise this problem in the same vein.

When we are explicitly told to move one of a series of independent objects, it is implicit that the other objects do not move or change - We don't need to be told.

Move the 9 from the end to the front

The wording of the problem has since changed, but I still like my number blocks.

Let the original number be $n$ . Then the LHS (left-hand side) of the equation below means moving the end 9 to the front and the RHS (right-hand side) is four times of the original number.

$\begin{aligned} \frac {n-9}{10} + 9\times 10^5 & = 4n \\ n-9 + 9000000 & = 40n \\ 39n & = 8999991 \\ \implies n & = \frac {8999991}{39} \\ & = \boxed{230769} \end{aligned}$

The solutions of H K and Chew-Seong Cheong are the most straightforward. I thought I'd pitch in with an alternative approach, of increasingly improved approximations.

First, write zeroes on the blanks on the right side. $???\,???\times 4 = 900\,000.$ Divide by four to determine what the factor on the left would be: $225\,000\times 4 = 900\,000.$ Copy the first five digits from the left onto the blanks. Divide by four again. $???\,???\times 4 = 922\,500.$ Again we divide by four: $230\,625\times 4 = 922\,500.$ Again copy the digits. I round off so that we get a multiple of four. $???\,???\times 4 = 923\,064.$ Thus: $230\,766\times 4 = 923\,064.$ And again: $???\,???\times 4 = 923\,076.$ It turns out that $230\,769\times 4 = 923\,076,$ so that we have the solution.

x= 100,000a + 10,000b + 1000c + 100d + 10e + 9

{x}=900,000 + 10,000a + 1000b + 100c + 10d + e

Make 10e the subject of the top equation: 10e = x - 100,000a - 10,000b - 1000c - 100d - 9 Divide both sides by 10.

e= $\frac{x - 100,000a - 10,000b - 1000c - 100d - 9}{10}$

e= $\frac{x}{10}$ -10,000a - 1000b - 100c - 10d - 0.9

As you can see, plugging this value of e into the second equation above conveniently cancels out all the other variables.

{x}=900,000 + 10,000a + 1000b + 100c + 10d + $\frac{x}{10}$ -10,000a - 1000b - 100c - 10d - 0.9

{x} = 899,999.1 + $\frac{x}{10}$

4x = 899,999.1 + $\frac{x}{10}$

$\frac{39}{10}$ x = 899,999.1

x = 899,999.1 / 39 * 10

x = 230,769

Proof of correctness- 923,076 / 4 = 230,679

Let $abcde9$ be the first digit and $9abcde$ the second digit.

The number $abcde9 \times 4 = (4a)(4b)(4c)(4d)(4e)(36)$

We have to solve the equation identifying each rank : $(4a)(4b)(4c)(4d)(4e)(36)$ is equal to $(9)(a)(b)(c)(d)(e)$

$(4a)(4b)(4c)(4d)(4e)(36) =(9)(a)(b)(c)(d)(e)$

Let's modify rank $10^1$ :

$(4e)({\color{#D61F06}36}) = (4e+{\color{#D61F06}3})({\color{#D61F06}6}) = (d)(e)$ so $\boxed{e=6}$ but ${\color{#3D99F6}(4e+3) = d}$ so $(d= 24 + 3 = 27)$

Let's modify rank $10^2$ :

$(4d)({\color{#D61F06}27})(6) = (4d+{\color{#D61F06}2)}({\color{#D61F06}7})(6)$ but ${\color{#3D99F6}(4d+2)=30}$ then $\boxed{d=7}$

Let's modify rank $10^3$ :

$(4c)({\color{#D61F06}30})(7)(6) = (4c+{\color{#D61F06}3})({\color{#D61F06}0})(7)(6)$ but ${\color{#3D99F6}(4c+3)=b}$ and $\boxed{c=0}$ then $\boxed{b=3}$

Let's modify rank $10^4$ :

$(4a)({\color{#D61F06}12})(3)(0)(7)(6) = (4a+{\color{#D61F06}1})({\color{#D61F06}2})(3)(0)(7)(6)$ but ${\color{#3D99F6}4b=a}$ so $(a=12)$

Let's modify rank $10^5$ :

$(4a)(12) = (4a+1)(2) = (9)(a)$ then we have ${\color{#3D99F6}(4a+1)=9}$ and ${\color{#3D99F6}a=2}$ then $\boxed{a=2}$

The original number required is $\boxed{{\color{#D61F06}230769}}$ such as $230769 \times 4 = 923076$

1|0|3|2|3| Remainder

2|3|0|7|6|9 Initial Figure

X|0|0|0|0|4 Multiplied by

9|2|3|0|7|6 Result

9x4=36, bring 6 down to the ones position in the result and 3 as the remainder, then bring the 6 up to the tens position in initial figure.

6x4=24+3=27, bring the 7 down to the tens position in the result and 2 as the remainder, then bring the 7 up to the hundreds position in initial figure.

7x4=28+2=30, bring the 0 down to the hundreds position in the result and 3 as the remainder, then bring the 0 up to the thousands position in the initial figure.

0x4=0+3=3, bring the 3 down to the thousands position in the result, then bring the 3 up to the ten thousands position in the initial figure.

3x4=12, bring the 2 down to the ten thousands position in the result and 1 as the remainder, then bring the 2 up to the hundred thousands position in the initial figure.

an iterative solution using a calculator: 900000/4=225000; 922500/4=230625; 923062/4=230766.25; 923076/4=230769; solved