if(x-10)(y-5)(z-2)=1000 then what is the least value of x+y+z? if x,y,z are all integer no.

• Answer :- 7
  (x-10)(y-5)(z-2)=1000
  put x=0
  y=-5
  z=12
  then
  (0-10)(-5-5)(12-2)=1000
  (-10)(-10)(10)=1000
  1000=1000
• 9 years agoHelpfull: Yes(32) No(1)
• Answer:47
  (x-10)(y-5)(z-2)=10^3
  that means x-10=10
  y-5=10
  z-2=10
  If we solve these three equations we can get x,y,z values as 20,15,12 respectively.Then
  x+y+z must be 47.
• 9 years agoHelpfull: Yes(8) No(7)
• Since we have to get least value,
  x=0
  y=-5
  z=12
  so x+y+z=7
• 9 years agoHelpfull: Yes(5) No(1)
• 1000=2^ 3*5^3
  1000=(35-10)(13-5)(7-2)
  then x+y+z= 35+13+7=55
• 9 years agoHelpfull: Yes(0) No(5)
• 1000= 10*10*10
  or (20-10)(15-5) (12-2)= 1000
  x=20 .y= 15,z=12 x+y+z= 47
• 9 years agoHelpfull: Yes(0) No(3)
• (x-10)=-10
  (y-5)=-5
  (z-2)=20 then x=0,y=0,z=22.Therefore,x+y+z=22
  
• 9 years agoHelpfull: Yes(0) No(4)
• Sneha paliwal @ what u will say if I show -983 as their sum.
  X=-990, Y=4, Z=3
  (-1000)(-1)(1)=1000.
  Source : fb
• 9 years agoHelpfull: Yes(0) No(5)
• 1000 can be written in factored form as 4*25*10
  so if (x-10)=4; x=14
  if (y-5)= 25; y=30
  if (z-2)=10; z=12
  14+30+12=56
  56 is the aanswer
• 9 years agoHelpfull: Yes(0) No(7)