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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

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2 Answers

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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 $6ABCDEF=857142$.

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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 $YX= DEFABC= 142857$.

$6(142857)= 857142$.

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