if 31x+30y+29z=366 then x+y+z=? where x, y, z are natural numbers.

• considering 12 months of a year
  31*7+30*4+29*1= 366
  
  so x=7,y=4,z=1
  x+y+z=12
• 11 years agoHelpfull: Yes(12) No(1)
• writing 31x+30y+29z = 30(x+y+z)+(x-y)=366 ,it is clear that x+y+z=12 and x-z=6 (which is acceptable here)as x+y+z=11 gives x-z=36 and x+y+z=13 gives z-x= 24 which is not possible here.
• 11 years agoHelpfull: Yes(4) No(3)
• 31(x + y + z) - ( y + 2z ) = 366
  x + y + z = (366 + y + 2z) / 31
  
  Where 366 + y + 2z is a multiple of 31, (y+2z) > 2 but < 7
  => x + y + z = 12
• 11 years agoHelpfull: Yes(2) No(2)
• (30+1)x + 30y + (30-1)z = 366
  So 30(x+y+z) + (x-z) = 366
  Now option a)30*9+(x-z) = 270 + (x-z)=366, so x-z = 96 which is not possible
  similarly for x+y+z =11
  So only option 12.
  As 30*12+(x-z) = 366 and 360+6 =366 is posible for x-z = 6.
  So ans = 12
  
• 9 years agoHelpfull: Yes(1) No(0)