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Question Number 202716 by Rasheed.Sindhi last updated on 01/Jan/24

abcd ^(−) is a four digit number such that a^2 +b^2 +c^2 +d^2 = cd ^(−) and cd ^(−) − d ^(−) = ab ^(−) . Find the number.

$$\overline {\:\:{abcd}\:\:}{is}\:{a}\:{four}\:{digit}\:{number} \\ $$$${such}\:{that}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\overline {\:{cd}\:} \\ $$$${and}\:\overline {\:{cd}\:}−\overline {\:{d}\:}=\overline {\:{ab}\:}. \\ $$$$\mathcal{F}{ind}\:{the}\:{number}. \\ $$

Answered by JDamian last updated on 01/Jan/24

cd^(__) − d^(_) = ab^(__) = c0^(__) ⇒ b=0 ∧ a=c a^2 +b^2 +c^2 +d^2 = cd^(__) =10c+d c^2 +0^2 +c^2 +d^2 = 10c+d 2c^2 +d^2 =10c+d d^2 −d = 10c−2c^2 d(d−1) = 2c(5−c) 2c(5−c)≥0 ⇒ c≤5 determinant ((c,(2c(5−c))),(0,0),(1,8),(2,(12)),(3,(12)),(4,8),(5,0)) determinant ((d,(d(d−1))),(1,0),(2,2),(3,6),(4,(12)),(5,(20)),(⋮,⋮)) d=4 ∧ c ∈ {2, 3} cd^(__) ∈ {24, 34} abcd^(____) ∈ {2024, 3034} Happy new year 2024... or 3034 if anybody reads this in 3034!!! lol

$$\overset{\_\_} {{cd}}\:−\:\overset{\_} {{d}}\:\:=\:\:\overset{\_\_} {{ab}}\:\:=\:\overset{\_\_} {{c}\mathrm{0}}\:\:\Rightarrow\:\:{b}=\mathrm{0}\:\:\:\wedge\:\:{a}={c} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\:\overset{\_\_} {{cd}}\:=\mathrm{10}{c}+{d} \\ $$$${c}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\:\mathrm{10}{c}+{d} \\ $$$$\mathrm{2}{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{10}{c}+{d} \\ $$$${d}^{\mathrm{2}} −{d}\:=\:\mathrm{10}{c}−\mathrm{2}{c}^{\mathrm{2}} \\ $$$$\boldsymbol{{d}}\left(\boldsymbol{{d}}−\mathrm{1}\right)\:=\:\mathrm{2}\boldsymbol{{c}}\left(\mathrm{5}−\boldsymbol{{c}}\right) \\ $$$$\mathrm{2}{c}\left(\mathrm{5}−{c}\right)\geqslant\mathrm{0}\:\:\Rightarrow\:{c}\leqslant\mathrm{5} \\ $$$$ \\ $$$$\begin{array}{|c|c|c|c|c|c|c|}{{c}}&\hline{\mathrm{2}{c}\left(\mathrm{5}−{c}\right)}\\{\mathrm{0}}&\hline{\mathrm{0}}\\{\mathrm{1}}&\hline{\mathrm{8}}\\{\mathrm{2}}&\hline{\mathrm{12}}\\{\mathrm{3}}&\hline{\mathrm{12}}\\{\mathrm{4}}&\hline{\mathrm{8}}\\{\mathrm{5}}&\hline{\mathrm{0}}\\\hline\end{array}\:\:\:\:\begin{array}{|c|c|c|c|c|c|c|}{{d}}&\hline{{d}\left({d}−\mathrm{1}\right)}\\{\mathrm{1}}&\hline{\mathrm{0}}\\{\mathrm{2}}&\hline{\mathrm{2}}\\{\mathrm{3}}&\hline{\mathrm{6}}\\{\mathrm{4}}&\hline{\mathrm{12}}\\{\mathrm{5}}&\hline{\mathrm{20}}\\{\vdots}&\hline{\vdots}\\\hline\end{array} \\ $$$$ \\ $$$${d}=\mathrm{4}\:\:\wedge\:\:{c}\:\in\:\left\{\mathrm{2},\:\mathrm{3}\right\} \\ $$$$ \\ $$$$\overset{\_\_} {{cd}}\:\in\:\left\{\mathrm{24},\:\mathrm{34}\right\} \\ $$$$ \\ $$$$\overset{\_\_\_\_} {{abcd}}\:\in\:\left\{\mathrm{2024},\:\mathrm{3034}\right\} \\ $$$$ \\ $$$$\boldsymbol{{Happy}}\:\boldsymbol{{new}}\:\boldsymbol{{year}}\:\mathrm{2024}...\:{or}\:\mathrm{3034}\:{if} \\ $$$${anybody}\:{reads}\:{this}\:{in}\:\mathrm{3034}!!!\:\:{lol} \\ $$

Commented by Rasheed.Sindhi last updated on 01/Jan/24

V. Nice!....Thanks sir! Happy new year 2024_(to you too) In advance, Happy year 3034 :)

$$\boldsymbol{\mathrm{V}}.\:\boldsymbol{\mathrm{N}}\mathrm{i}\boldsymbol{\mathrm{ce}}!....\mathcal{T}\boldsymbol{{hanks}}\:\boldsymbol{{sir}}! \\ $$$$\underset{\boldsymbol{{to}}\:\boldsymbol{{you}}\:\boldsymbol{{too}}} {\boldsymbol{{Happy}}\:\boldsymbol{{new}}\:\boldsymbol{{year}}\:\mathrm{2024}}\: \\ $$$$\boldsymbol{{In}}\:\boldsymbol{{advance}},\: \\ $$$$\left.\boldsymbol{{Happy}}\:\boldsymbol{{year}}\:\mathrm{3034}\::\right) \\ $$

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