Find an integral solution of the expression
31x + 28y + 30z = 365

(P.S.: I was given 45 seconds to think)

• x = 7 , y = 1 , z = 4
  
  a year of 365 days,
  7 months of 31 days, 1 of 28 days, 4 of 30 days .
• 9 years agoHelpfull: Yes(3) No(0)
• days of a year=365
  x=7 i.e. 7 months having 31 days
  y=1 i.e. 1 month have 28 days
  z=4 i.e. 4 months having 30 days
• 9 years agoHelpfull: Yes(1) No(0)
• x=9--->279
  y=2----> 56
  z=1----> 30
  -------------------
  365
  --------------------
• 9 years agoHelpfull: Yes(0) No(0)
• x+y+z=12
  no of moths having 31 days = }
  no of months having 28 days = } =365 days
  no of months having 30 days =}
  therefore 12 months=365 days
• 9 years agoHelpfull: Yes(0) No(0)
• 31*5+28*0+30*7
• 4 years agoHelpfull: Yes(0) No(0)