28a + 30 b +31c =365
find a+b+c if a.b.c are natural number

• 28a+30b+31c =365.a,b,c are in the form of months
  means a=feb=1,b=4,c=7,means 28*1+30*4+31*7=365
  thn a+b+c=1+4+7=12
• 11 years agoHelpfull: Yes(21) No(5)
• This is an instance of Calendar Problem.
  a = #months with total days = 28.
  b = #months with total days = 30.
  c = #months with total days = 31.
  
  a+b+c = Total Months in an year = 12
• 11 years agoHelpfull: Yes(13) No(1)
• If you observe carefully 28,30and 31 and no of days in the months of the year, which sums up to 365. so a,b,c represent number of days with 28,30 and 31 days in it so a=1,b=4,c=7 and a+b+c=12
• 11 years agoHelpfull: Yes(1) No(0)
• 12 is the ans
  
• 11 years agoHelpfull: Yes(0) No(0)
• a=2,b=1,c=9
  i.e. a+b+c=12
• 11 years agoHelpfull: Yes(0) No(4)
• 28(1)+30(4)+31(7)=365
  this are days in every months of the year..
• 11 years agoHelpfull: Yes(0) No(0)
• a is feb month and hense a=1
  b is 30 days months hense b=4
  c is 31 days months hense c=7
  28*1+30*4+31*7=365
  1+4+7=12
• 11 years agoHelpfull: Yes(0) No(0)
• 12 a,b,c are months of year
• 11 years agoHelpfull: Yes(0) No(0)
• its very simple.
  28(1)+30(4)+31(7)=365(a year)
  s0 1+4+7=12
  which means 12 months in a year
  ans is 12
  
• 11 years agoHelpfull: Yes(0) No(0)