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Question Number 114130 by bemath last updated on 17/Sep/20

Commented by mr W last updated on 17/Sep/20

z=2: x2^(−) ×4y2^(−) =7344 (10x+2)(400+10y+2)=7344 4020x+804+100xy+20y=7344 ⇒x=1 120y=2520 ⇒y=21>9 ⇒no solution z=8: x8^(−) ×4y8^(−) =7344 (10x+8)(408+10y)=7344 4080x+100xy+80y+408×8=7344 4080x+100xy+80y=4080 ⇒x=1 180y=0 ⇒y=0 ⇒x=1, y=0, z=8 x+y+z=9 check: 18×408=7344

$${z}=\mathrm{2}: \\ $$$$\overline {{x}\mathrm{2}}×\overline {\mathrm{4}{y}\mathrm{2}}=\mathrm{7344} \\ $$$$\left(\mathrm{10}{x}+\mathrm{2}\right)\left(\mathrm{400}+\mathrm{10}{y}+\mathrm{2}\right)=\mathrm{7344} \\ $$$$\mathrm{4020}{x}+\mathrm{804}+\mathrm{100}{xy}+\mathrm{20}{y}=\mathrm{7344} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$$\mathrm{120}{y}=\mathrm{2520} \\ $$$$\Rightarrow{y}=\mathrm{21}>\mathrm{9}\:\Rightarrow{no}\:{solution} \\ $$$$ \\ $$$${z}=\mathrm{8}: \\ $$$$\overline {{x}\mathrm{8}}×\overline {\mathrm{4}{y}\mathrm{8}}=\mathrm{7344} \\ $$$$\left(\mathrm{10}{x}+\mathrm{8}\right)\left(\mathrm{408}+\mathrm{10}{y}\right)=\mathrm{7344} \\ $$$$\mathrm{4080}{x}+\mathrm{100}{xy}+\mathrm{80}{y}+\mathrm{408}×\mathrm{8}=\mathrm{7344} \\ $$$$\mathrm{4080}{x}+\mathrm{100}{xy}+\mathrm{80}{y}=\mathrm{4080} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$$\mathrm{180}{y}=\mathrm{0} \\ $$$$\Rightarrow{y}=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{x}=\mathrm{1},\:{y}=\mathrm{0},\:{z}=\mathrm{8} \\ $$$${x}+{y}+{z}=\mathrm{9} \\ $$$$ \\ $$$${check}: \\ $$$$\mathrm{18}×\mathrm{408}=\mathrm{7344} \\ $$

Commented by bemath last updated on 17/Sep/20

thank you prof

$${thank}\:{you}\:{prof} \\ $$

Commented by mr W last updated on 17/Sep/20

santuyy!

$${santuyy}! \\ $$

Commented by bemath last updated on 17/Sep/20

wakwakwak....

$${wakwakwak}.... \\ $$

Commented by Rasheed.Sindhi last updated on 17/Sep/20

Excellent Sir! Perhaps the nicest!

$$\mathcal{E}{xcellent}\:\mathcal{S}{ir}! \\ $$$${Perhaps}\:{the}\:{nicest}! \\ $$

Answered by 1549442205PVT last updated on 17/Sep/20

xz^(−) .4yz^(−) =7344⇔(10x+z)(400+10y+z)=7344 ⇔(4000x+400z+100xy+10yz+10xz) +z^2 =7344(1)⇔(7344−z^2 )⋮10⇒z∈{2,8} i)If z=2 then replace into (1)we get 4000x+800+100xy+20y+20x+4=7344 ⇔4020x+100xy+20y=6540 ⇔402x+10xy+2y=654 ⇔201x+5xy+y=327⇒y=((327−201x)/(5x+1)) ⇒5y=((327.5−5.201x)/(5x+1))=−201+((1836)/(5x+1)) ⇒(5y+201)(5x+1)=1836 ⇒5y+201∣1836=3^3 .2^2 .17 Since 0≤ y≤9,201<5y+201≤246 The divisors biger 201 of 1836 are 204,306, ⇒5y+201=204=17.12⇒y∉N ii)If z=8 then replace into (1)we get 4000x+3200+100xy+80y+80x+64=7344 ⇔4080x+100xy+80y=4080 ⇔408x+10xy+8y=408 ⇔204x+5xy+4y=204(2) Since 5xy+4y≥0 and 204x≥204,so (2)⇔ { ((5xy+4y=0)),((204x=204)) :}⇔ { ((y=0)),((x=1)) :} Thus,our problem has unique solution (x,y,z)=(1,0,8)⇒x+y+z=9

$$\overline {\mathrm{xz}}.\mathrm{4}\overline {\mathrm{yz}}=\mathrm{7344}\Leftrightarrow\left(\mathrm{10x}+\mathrm{z}\right)\left(\mathrm{400}+\mathrm{10y}+\mathrm{z}\right)=\mathrm{7344} \\ $$$$\Leftrightarrow\left(\mathrm{4000x}+\mathrm{400z}+\mathrm{100xy}+\mathrm{10yz}+\mathrm{10xz}\right) \\ $$$$+\mathrm{z}^{\mathrm{2}} =\mathrm{7344}\left(\mathrm{1}\right)\Leftrightarrow\left(\mathrm{7344}−\mathrm{z}^{\mathrm{2}} \right)\vdots\mathrm{10}\Rightarrow\mathrm{z}\in\left\{\mathrm{2},\mathrm{8}\right\} \\ $$$$\left.\mathrm{i}\right)\mathrm{If}\:\mathrm{z}=\mathrm{2}\:\mathrm{then}\:\mathrm{replace}\:\mathrm{into}\:\left(\mathrm{1}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{4000x}+\mathrm{800}+\mathrm{100xy}+\mathrm{20y}+\mathrm{20x}+\mathrm{4}=\mathrm{7344} \\ $$$$\Leftrightarrow\mathrm{4020x}+\mathrm{100xy}+\mathrm{20y}=\mathrm{6540} \\ $$$$\Leftrightarrow\mathrm{402x}+\mathrm{10xy}+\mathrm{2y}=\mathrm{654} \\ $$$$\Leftrightarrow\mathrm{201x}+\mathrm{5xy}+\mathrm{y}=\mathrm{327}\Rightarrow\mathrm{y}=\frac{\mathrm{327}−\mathrm{201x}}{\mathrm{5x}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{5y}=\frac{\mathrm{327}.\mathrm{5}−\mathrm{5}.\mathrm{201x}}{\mathrm{5x}+\mathrm{1}}=−\mathrm{201}+\frac{\mathrm{1836}}{\mathrm{5x}+\mathrm{1}} \\ $$$$\Rightarrow\left(\mathrm{5y}+\mathrm{201}\right)\left(\mathrm{5x}+\mathrm{1}\right)=\mathrm{1836} \\ $$$$\Rightarrow\mathrm{5y}+\mathrm{201}\mid\mathrm{1836}=\mathrm{3}^{\mathrm{3}} .\mathrm{2}^{\mathrm{2}} .\mathrm{17} \\ $$$$\mathrm{Since}\:\mathrm{0}\leqslant\:\mathrm{y}\leqslant\mathrm{9},\mathrm{201}<\mathrm{5y}+\mathrm{201}\leqslant\mathrm{246} \\ $$$$\mathrm{The}\:\mathrm{divisors}\:\mathrm{biger}\:\mathrm{201}\:\mathrm{of}\:\mathrm{1836}\:\mathrm{are} \\ $$$$\mathrm{204},\mathrm{306}, \\ $$$$\Rightarrow\mathrm{5y}+\mathrm{201}=\mathrm{204}=\mathrm{17}.\mathrm{12}\Rightarrow\mathrm{y}\notin\mathrm{N} \\ $$$$\left.\mathrm{ii}\right)\mathrm{If}\:\mathrm{z}=\mathrm{8}\:\mathrm{then}\:\mathrm{replace}\:\mathrm{into}\:\left(\mathrm{1}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{4000x}+\mathrm{3200}+\mathrm{100xy}+\mathrm{80y}+\mathrm{80x}+\mathrm{64}=\mathrm{7344} \\ $$$$\Leftrightarrow\mathrm{4080x}+\mathrm{100xy}+\mathrm{80y}=\mathrm{4080} \\ $$$$\Leftrightarrow\mathrm{408x}+\mathrm{10xy}+\mathrm{8y}=\mathrm{408} \\ $$$$\Leftrightarrow\mathrm{204x}+\mathrm{5xy}+\mathrm{4y}=\mathrm{204}\left(\mathrm{2}\right) \\ $$$$\mathrm{Since}\:\mathrm{5xy}+\mathrm{4y}\geqslant\mathrm{0}\:\mathrm{and}\:\mathrm{204x}\geqslant\mathrm{204},\mathrm{so} \\ $$$$\left(\mathrm{2}\right)\Leftrightarrow\begin{cases}{\mathrm{5xy}+\mathrm{4y}=\mathrm{0}}\\{\mathrm{204x}=\mathrm{204}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{y}=\mathrm{0}}\\{\mathrm{x}=\mathrm{1}}\end{cases} \\ $$$$\mathrm{Thus},\mathrm{our}\:\mathrm{problem}\:\mathrm{has}\:\mathrm{unique}\:\mathrm{solution} \\ $$$$\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right)=\left(\mathrm{1},\mathrm{0},\mathrm{8}\right)\Rightarrow\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}=\mathrm{9} \\ $$$$ \\ $$

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