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abac-dc-2-d-ab-c-c-2-ac-abac-




Question Number 215272 by Rasheed.Sindhi last updated on 02/Jan/25
 { ((abac^(−) =(dc^(−) )^2 )),((d=((ab^(−) )/c))),((c^2 =ac^(−) )) :}  abac^(−) =?
$$\begin{cases}{\overline {{abac}}=\left(\overline {{dc}}\right)^{\mathrm{2}} }\\{{d}=\frac{\overline {{ab}}}{{c}}}\\{{c}^{\mathrm{2}} =\overline {{ac}}}\end{cases} \\ $$$$\overline {{abac}}=? \\ $$
Answered by A5T last updated on 02/Jan/25
c^2 =ac^(−) ⇒c=0,1, 5 or 6    ;   d=((ab^(−) )/c)⇒c≠0  c=1⇒a=0⇒d=b  ⇒100b+1=(10d+1)^2 ⇒100b+1=100b^2 +1+20b  ⇒100b^2 −80b=0⇒b=0⇒d=0  a=b=d=0;c=1⇒abac^(−) =0001    c=5⇒a=2⇒5d=20+b⇒(d,b)=(4,0);(5,5)  Checking⇒(d,b)=(4,0)⇒abac^(−) =2025    c=6⇒a=3⇒6d=30+b⇒(d,b)=(5,0),(6,6)  Checking⇒ None works    ⇒abac^(−) =2025 or 0001
$$\mathrm{c}^{\mathrm{2}} =\overline {\mathrm{ac}}\Rightarrow\mathrm{c}=\mathrm{0},\mathrm{1},\:\mathrm{5}\:\mathrm{or}\:\mathrm{6}\:\:\:\:;\:\:\:\mathrm{d}=\frac{\overline {\mathrm{ab}}}{\mathrm{c}}\Rightarrow\mathrm{c}\neq\mathrm{0} \\ $$$$\mathrm{c}=\mathrm{1}\Rightarrow\mathrm{a}=\mathrm{0}\Rightarrow\mathrm{d}=\mathrm{b} \\ $$$$\Rightarrow\mathrm{100b}+\mathrm{1}=\left(\mathrm{10d}+\mathrm{1}\right)^{\mathrm{2}} \Rightarrow\mathrm{100b}+\mathrm{1}=\mathrm{100b}^{\mathrm{2}} +\mathrm{1}+\mathrm{20b} \\ $$$$\Rightarrow\mathrm{100b}^{\mathrm{2}} −\mathrm{80b}=\mathrm{0}\Rightarrow\mathrm{b}=\mathrm{0}\Rightarrow\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{a}=\mathrm{b}=\mathrm{d}=\mathrm{0};\mathrm{c}=\mathrm{1}\Rightarrow\overline {\mathrm{abac}}=\mathrm{0001} \\ $$$$ \\ $$$$\mathrm{c}=\mathrm{5}\Rightarrow\mathrm{a}=\mathrm{2}\Rightarrow\mathrm{5d}=\mathrm{20}+\mathrm{b}\Rightarrow\left(\mathrm{d},\mathrm{b}\right)=\left(\mathrm{4},\mathrm{0}\right);\left(\mathrm{5},\mathrm{5}\right) \\ $$$$\mathrm{Checking}\Rightarrow\left(\mathrm{d},\mathrm{b}\right)=\left(\mathrm{4},\mathrm{0}\right)\Rightarrow\overline {\mathrm{abac}}=\mathrm{2025} \\ $$$$ \\ $$$$\mathrm{c}=\mathrm{6}\Rightarrow\mathrm{a}=\mathrm{3}\Rightarrow\mathrm{6d}=\mathrm{30}+\mathrm{b}\Rightarrow\left(\mathrm{d},\mathrm{b}\right)=\left(\mathrm{5},\mathrm{0}\right),\left(\mathrm{6},\mathrm{6}\right) \\ $$$$\mathrm{Checking}\Rightarrow\:\mathrm{None}\:\mathrm{works} \\ $$$$ \\ $$$$\Rightarrow\overline {\mathrm{abac}}=\mathrm{2025}\:\mathrm{or}\:\mathrm{0001} \\ $$
Commented by Rasheed.Sindhi last updated on 02/Jan/25
Happy 2025 sir!  Thanks for nice answer!
$$\mathcal{H}{appy}\:\mathrm{2025}\:{sir}! \\ $$$$\mathcal{T}{hanks}\:{for}\:{nice}\:{answer}! \\ $$
Commented by A5T last updated on 02/Jan/25
Same to you, thanks.
$$\mathrm{Same}\:\mathrm{to}\:\mathrm{you},\:\mathrm{thanks}. \\ $$

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