Find all Ω=abcdef ^(−) , such that abcdef=abc+def


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Question Number 196672 by Rasheed.Sindhi last updated on 29/Aug/23

$F i n d a l l Ω \overset{―}{= a b c d e f}, s u c h t h a t$ $a b c d e f = a b c + d e f$
Commented by Rasheed.Sindhi last updated on 29/Aug/23

$You can't use 'macro parameter character #' in math mode$
	Answered by Rasheed.Sindhi last updated on 30/Aug/23
	$abcdef=abc+def abcdef−abc=def abc(def−1)=def abc=((def)/(def−1)) ⇒ (def−1)∣ def⇒def−1=1 ⇒def=2 { ((d=2,e=1,f=1)),((d=1,e=2,f=1)),((d=1,e=1,f=2)) :} def=211,121,112 abc=((def)/(def−1))=(2/1)=2 abc=211,121,112 Ω=abcdef^(−) =211211,211121,211112, 121211,121121,121112, 112211,112121,112112$
	$a b c d e f = a b c + d e f$ $a b c d e f - a b c = d e f$ $a b c (d e f - 1) = d e f$ $a b c = \frac{d e f}{d e f - 1} \Rightarrow (d e f - 1) ∣ d e f \Rightarrow d e f - 1 = 1$ $\Rightarrow d e f = 2$ ${\begin{cases} d = 2, e = 1, f = 1 \\ d = 1, e = 2, f = 1 \\ d = 1, e = 1, f = 2 \end{cases}$ $d e f = 211, 121, 112$ $a b c = \frac{d e f}{d e f - 1} = \frac{2}{1} = 2$ $a b c = 211, 121, 112$ $Ω = \overset{―}{a b c d e f}$ $= 211211, 211121, 211112,$ $121211, 121121, 121112,$ $112211, 112121, 112112$
	Commented by Tawa11 last updated on 29/Aug/23

	$Long time sir .$
	Commented by Rasheed.Sindhi last updated on 30/Aug/23

	$Y e s m i s s, n o w a d a y s I r a r e l y c a n$ $s h a r e a n y p o s t s a n d y o u a l s o l e s s$ $p a r t i c i p a t e!$ $A n y w a y t h a n k s m i s s .$
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