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

Commented bymr 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

Commented bybemath last updated on 17/Sep/20

thank you prof

Commented bymr W last updated on 17/Sep/20

santuyy!

Commented bybemath last updated on 17/Sep/20

wakwakwak....

Commented byRasheed.Sindhi last updated on 17/Sep/20

Excellent Sir!  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