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

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

$${Find}\:{all}\:\Omega\overline {={abcdef}\:\:},\:{such}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{abcdef}={abc}+{def} \\ $$

Commented by Rasheed.Sindhi last updated on 29/Aug/23

Q#196659

$${Q}#\mathrm{196659} \\ $$

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

$${abcdef}={abc}+{def} \\ $$$${abcdef}−{abc}={def} \\ $$$${abc}\left({def}−\mathrm{1}\right)={def} \\ $$$${abc}=\frac{{def}}{{def}−\mathrm{1}}\:\:\:\:\Rightarrow\:\left({def}−\mathrm{1}\right)\mid\:{def}\Rightarrow{def}−\mathrm{1}=\mathrm{1} \\ $$$$\Rightarrow{def}=\mathrm{2} \\ $$$$\begin{cases}{{d}=\mathrm{2},{e}=\mathrm{1},{f}=\mathrm{1}}\\{{d}=\mathrm{1},{e}=\mathrm{2},{f}=\mathrm{1}}\\{{d}=\mathrm{1},{e}=\mathrm{1},{f}=\mathrm{2}}\end{cases} \\ $$$${def}=\mathrm{211},\mathrm{121},\mathrm{112} \\ $$$${abc}=\frac{{def}}{{def}−\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{1}}=\mathrm{2} \\ $$$${abc}=\mathrm{211},\mathrm{121},\mathrm{112} \\ $$$$\Omega=\overline {{abcdef}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{211211},\mathrm{211121},\mathrm{211112}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{121211},\mathrm{121121},\mathrm{121112}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{112211},\mathrm{112121},\mathrm{112112} \\ $$

Commented by Tawa11 last updated on 29/Aug/23

Long time sir.

$$\mathrm{Long}\:\mathrm{time}\:\mathrm{sir}. \\ $$

Commented by Rasheed.Sindhi last updated on 30/Aug/23

Yes miss, nowadays I rarely can share any posts and you also less participate! Anyway thanks miss.

$${Yes}\:{miss},\:{nowadays}\:{I}\:{rarely}\:{can} \\ $$$$\:{share}\:{any}\:{posts}\:{and}\:{you}\:{also}\:{less} \\ $$$$\:{participate}! \\ $$$${Anyway}\:{thanks}\:{miss}. \\ $$

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