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_ _ _ _ _ $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).

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    $\begingroup$ Aren't $E$ and $e$ the same? $\endgroup$ – Michael Burr May 6 '17 at 11:39
  • $\begingroup$ ^Exactly my thought $\endgroup$ – Kugelblitz May 6 '17 at 11:40
  • $\begingroup$ @MichaelBurr No, why? See the first line after the quotation carefully. $\endgroup$ – Soha Farhin Pine May 6 '17 at 11:40
  • $\begingroup$ The way that the problem is described, it seems like all the digits stay the same, just the $9$ in the units digit is moved to the front of the number. Using the set-up you describe, I doubt that the problem has a unique solution since you're asking which $6$ digit numbers ending in $9$ times $4$ become $6$ digit numbers beginning with $9$. There are many of these. $\endgroup$ – Michael Burr May 6 '17 at 11:42
  • $\begingroup$ But, the source of this problem clearly meant that there is only one solution. $\endgroup$ – Soha Farhin Pine May 6 '17 at 11:46
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I am interpreting this question in a different way than the OP. The question is about the operation being performed. The question states that you start with a $6$ digit number which ends in a $9$, say $$ 123459 $$ and move the last digit (the $9$) to the front, you have a new $6$ digit number, in this case $$ 912345. $$ Observe that the other $5$ digits are unchanged, just their position changes. This matches the interpretation of @SimplyBeautifulArt. In the OP's question, the digits (other than $9$) can be anything, and there will not be a unique answer unless the structure between the other digits is used.

Now, this example doesn't satisfy the given conditions because it doesn't have the multiplication by $4$ property. Now, this problem can be solved directly. You know that $$ 4\cdot abcde9=9abcde $$ Since the units digit of the product on the left is $6$, $e=6$. Therefore, we now have $$ 4\cdot abcd69=9abcd6. $$ The $10$s digit on the left is $7$, so $d=7$. Hence, we have $$ 4\cdot abc769=9abc76. $$ Continuing wit the $100$s digit, we get that $c=0$. Therefore, we now have $$ 4\cdot ab0769=9ab076. $$ For the $1,000$s digit, we get that $b=3$, so we now have $$ 4\cdot a30769=9a3076. $$ Finally, the $10,000$s digit on the LHS is $2$, so $a=2$. Thus, we have $$ 4\cdot 230769=923076. $$ Therefore, the original number is $230769$.

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    $\begingroup$ Did it in the way OP was trying to. Nice. +1 $\endgroup$ – Kugelblitz May 6 '17 at 11:54
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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

$x=230769$

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  • $\begingroup$ For those unfamiliar with hidden lines, just hover over the end to see the solution. $\endgroup$ – Simply Beautiful Art May 6 '17 at 11:42
  • $\begingroup$ It can solved with an equation? $\endgroup$ – Soha Farhin Pine May 6 '17 at 11:44
  • $\begingroup$ Yes, an algebraic equation. Though the numbers are a tad bit large. $\endgroup$ – Simply Beautiful Art May 6 '17 at 11:45
  • $\begingroup$ Can it be solved the way I tried to? $\endgroup$ – Soha Farhin Pine May 6 '17 at 11:48
  • $\begingroup$ @SohaFarhinPine See MichaelBurr's answer. $\endgroup$ – Simply Beautiful Art May 6 '17 at 11:49
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Assuming $ABCDE$ is $abcde$ is the case: $$4[(abcde)\cdot10+9]=9\cdot10^5+abcde$$ That is, $$39(abcde)=9\cdot(10^5-4) => abcde=23076$$

So original is simply $230769$.


Note: Your method is too time consuming, and if only not possible to use equations, must you resort to trial and error.

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It says that if you move the last digit, which is a 9, to the front, then the number gets quadrupled.

So, you have a 6-digit number ABCDE9, so if you move the 9 to the front, you get 9ABCDE. We are told that this new number 9ABCDE is 4 times the origial number, so you need to solve:

ABCDE9 x 4 = 9ABCDE

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  • $\begingroup$ Thanks for clarifying the wording of the problem. $\endgroup$ – Soha Farhin Pine May 6 '17 at 12:05
  • $\begingroup$ @SohaFarhinPine You're welcome! :) $\endgroup$ – Bram28 May 6 '17 at 12:30
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So, the original number is in the form of $\overline{abcde9}$ and the second one is $\overline{9abcde}$. Therefore, you can "translate" it as: $$ 4\cdot\overline {abcde9}=\overline {9abcde}\tag1 $$

Now, let $x = \overline{abcde}$: $$ \overline {abcde9} = 10x + 9 $$ $$ \overline {9abcde} = 9 \cdot 10^5 + x $$

From (1): $$ 4\cdot(10x + 9) = 9 \cdot 10^5 + x $$ $$ 40 x + 36 = 9\cdot 10^5+x $$ $$ 39x = 899964 \implies x = 23076 $$

The original number is $\overline{x9}=230769$

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$\overline{abcde9}= 10^5a + 10^4b + 10^3c + 10^2d + 10e + 9$

$\overline{9abcde}= 9\times10^5 + 10^4a + 10^3b + 10^2c + 10d + e$

Let $x=\overline{abcde9}$. Making $10e$ the subject of the top equation:

10e = x - 100,000a - 10,000b - 1000c - 100d - 9 Divide both sides by 10.

e=

e= -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 + -10,000a - 1000b - 100c - 10d - 0.9

{x} = 899,999.1 +

4x = 899,999.1 +

x = 899,999.1

x = 899,999.1 / 39 * 10

x = 230,769

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