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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

$z = 2 :$ $\overset{―}{x 2} \times \overset{―}{4 y 2} = 7344$ $(10 x + 2) (400 + 10 y + 2) = 7344$ $4020 x + 804 + 100 x y + 20 y = 7344$ $\Rightarrow x = 1$ $120 y = 2520$ $\Rightarrow y = 21 > 9 \Rightarrow n o s o l u t i o n$ $z = 8 :$ $\overset{―}{x 8} \times \overset{―}{4 y 8} = 7344$ $(10 x + 8) (408 + 10 y) = 7344$ $4080 x + 100 x y + 80 y + 408 \times 8 = 7344$ $4080 x + 100 x y + 80 y = 4080$ $\Rightarrow x = 1$ $180 y = 0$ $\Rightarrow y = 0$ $\Rightarrow x = 1, y = 0, z = 8$ $x + y + z = 9$ $c h e c k :$ $18 \times 408 = 7344$
Commented by bemath last updated on 17/Sep/20

$t h a n k y o u p r o f$
Commented by mr W last updated on 17/Sep/20

$s a n t u y y!$
Commented by bemath last updated on 17/Sep/20

$w a k w a k w a k . . . .$
Commented by Rasheed.Sindhi last updated on 17/Sep/20

$E x c e l l e n t S i r!$ $P e r h a p s t h e n i c e s t!$
	Answered by 1549442205PVT last updated on 17/Sep/20
	$xz^(−) .4yz^(−) =7344⇔(10x+z)(400+10y+z)=7344 ⇔(4000x+400z+100xy+10yz+10xz) +z^2 =7344(1)⇔(7344−z^2 )⋮10⇒z∈{2,8} i)If z=2 then replace into (1)we get 4000x+800+100xy+20y+20x+4=7344 ⇔4020x+100xy+20y=6540 ⇔402x+10xy+2y=654 ⇔201x+5xy+y=327⇒y=((327−201x)/(5x+1)) ⇒5y=((327.5−5.201x)/(5x+1))=−201+((1836)/(5x+1)) ⇒(5y+201)(5x+1)=1836 ⇒5y+201∣1836=3^3 .2^2 .17 Since 0≤ y≤9,201<5y+201≤246 The divisors biger 201 of 1836 are 204,306, ⇒5y+201=204=17.12⇒y∉N ii)If z=8 then replace into (1)we get 4000x+3200+100xy+80y+80x+64=7344 ⇔4080x+100xy+80y=4080 ⇔408x+10xy+8y=408 ⇔204x+5xy+4y=204(2) Since 5xy+4y≥0 and 204x≥204,so (2)⇔ { ((5xy+4y=0)),((204x=204)) :}⇔ { ((y=0)),((x=1)) :} Thus,our problem has unique solution (x,y,z)=(1,0,8)⇒x+y+z=9$
	$\overset{―}{xz} . 4 \overset{―}{yz} = 7344 \Leftrightarrow (10 x + z) (400 + 10 y + z) = 7344$ $\Leftrightarrow (4000 x + 400 z + 100 xy + 10 yz + 10 xz)$ $+ z^{2} = 7344 (1) \Leftrightarrow (7344 - z^{2}) ⋮ 10 \Rightarrow z \in {2, 8}$ $i) If z = 2 then replace into (1) we get$ $4000 x + 800 + 100 xy + 20 y + 20 x + 4 = 7344$ $\Leftrightarrow 4020 x + 100 xy + 20 y = 6540$ $\Leftrightarrow 402 x + 10 xy + 2 y = 654$ $\Leftrightarrow 201 x + 5 xy + y = 327 \Rightarrow y = \frac{327 - 201 x}{5 x + 1}$ $\Rightarrow 5 y = \frac{327.5 - 5 . 201 x}{5 x + 1} = - 201 + \frac{1836}{5 x + 1}$ $\Rightarrow (5 y + 201) (5 x + 1) = 1836$ $\Rightarrow 5 y + 201 ∣ 1836 = 3^{3} . 2^{2} . 17$ $Since 0 ⩽ y ⩽ 9, 201 < 5 y + 201 ⩽ 246$ $The divisors biger 201 of 1836 are$ $204, 306,$ $\Rightarrow 5 y + 201 = 204 = 17.12 \Rightarrow y \notin N$ $ii) If z = 8 then replace into (1) we get$ $4000 x + 3200 + 100 xy + 80 y + 80 x + 64 = 7344$ $\Leftrightarrow 4080 x + 100 xy + 80 y = 4080$ $\Leftrightarrow 408 x + 10 xy + 8 y = 408$ $\Leftrightarrow 204 x + 5 xy + 4 y = 204 (2)$ $Since 5 xy + 4 y ⩾ 0 and 204 x ⩾ 204, so$ $(2) \Leftrightarrow {\begin{cases} 5 xy + 4 y = 0 \\ 204 x = 204 \end{cases} \Leftrightarrow {\begin{cases} y = 0 \\ x = 1 \end{cases}$ $Thus, our problem has unique solution$ $(x, y, z) = (1, 0, 8) \Rightarrow x + y + z = 9$
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