$6$ digit number evaluation satisfy the property $6(ABCDEF) = DEFABC$

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 $6(ABCDEF) = (DEFABC)$

$600000A+60000B+6000C+600D+60E+6F=100000D+10000E+1000F+100A+10B+C$

$599900A+59990B+5999C=99400D+9940E+994F$

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

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

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

$6(142857)= 857142$.