A $6$ digit number $ABCDEF$ when multiplied by $6$ gives the $6$ Digit number $DEFABC,$

then finding sum of digits of the number $DEFABC$ is

given $6(ABCDEF) = (DEFABC)$



$5999(100A+10B+C) = 994(100D+10E+F)$

I want to go further could some help me with this, Thanks

  • 2
    $\begingroup$ $5999 = 7 \cdot 857$ and $994 = 2 \cdot 7 \cdot 71$. Use this to help you solve $\endgroup$ – Benson Lin Oct 18 '16 at 12:34
  • 1
    $\begingroup$ Alternatively $\frac{1}{7}= 0.\overline{142857}$ $\endgroup$ – Henry Oct 18 '16 at 12:37
  • $\begingroup$ Is $6(ABCDEF)=DEFABC$ (as in the title) or is it $FEDCBA$ (as in the question)? Please clarify. $\endgroup$ – Parcly Taxel Oct 18 '16 at 12:43
  • $\begingroup$ parcy Taxel i have edited it. $\endgroup$ – DXT Oct 18 '16 at 12:48

By simplifying your equation, you arrive at

$$ 857(100A+10B+C) = 2\times71(100D+10E+F), $$

then you can conclude that in order for this equivalence to be true, also the following equations must be true: $$ \begin{align*} 100A+10B+C &= 2\times71\times K,\\ 100D+10E+F&=857\times K. \end{align*} $$

If you take $K=1$ you arrive at the solution: $$ \begin{align*} 100A+10B+C&=142,\\ 100D+10E+F&=857. \end{align*} $$

So $ABCDEF=142857$, and $6ABCDEF=857142$.


This can be more easily done as $XY= 6YX$ where $X$ and $Y$ are three digit numbers: $X= ABC$ and $Y= DEF$.

$1000X+ Y = 6(1000Y+ X)$.

Then $1000X+ Y= 6000Y+ 6X$ so $994X= 5999Y$.

$X= (5999/994)Y= (857/142)Y$ and since $X$ and $Y$ are three digit integers, we just take $Y= 142$ and $X= 857$.

$XY= ABCDEF= 857142$ and $YX= DEFABC= 142857$.

$6(142857)= 857142$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.