_ _ _ _ _ $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.


How can I possibly assume that the digits which fill the blanks of one side are identical to those of the other side?


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).

  • 1
    $\begingroup$ 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

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 $230769$.

  • 2
    $\begingroup$ Did it in the way OP was trying to. Nice. +1 $\endgroup$ – Kugelblitz May 6 '17 at 11:54


$$\underbrace{x=abcde9}_{\text{we want to solve for this}}$$

$$\underbrace{900000+abcde=4x}_{\text{this is moving the 9 to the front}}$$

$$\underbrace{abcde0=x-9}_{\text{this is dropping the 9 from the end}}$$

$$\underbrace{abcde=\frac1{10}abcde0}_{\text{this is moving the digits over}}$$



Solving this gives


  • $\begingroup$ 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

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.


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:


  • $\begingroup$ 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

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 $\overline{x9}=230769$


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

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

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= -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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