19.If f(x) = sum of all the digits of x, where x is a natural number, then what is the value of f(101)+f(102)+f(103)+ .. +f(200)?
Option 1 : 1000 Option 2 : 784 Option 3 : 999 Option 4 : 1001

• adding 99(100's places)+10(1+2+3+4+5+6+7+8+9) for 10's places +10(1+2+3+4+5+6+7+8+9)for unit places +2 (for 200)
  =99+450+450+2
  =1001
• 7 years agoHelpfull: Yes(8) No(1)
• if we add all digits from 101 to 200 the answer will be 1001
  from 101 to 110 = 2+3+4+5+6+7+8+9+10+2=56
  similarly from 111 to 120 =66
  when we add digits from 191 to 200 the total will be 137
  56+76+86+96+106+116+126+136+137=1001
• 7 years agoHelpfull: Yes(4) No(2)
• IF WE ADD all the digits from 101to 200 ans will be1001
  as us (2+3+4+5+6+7+8+9+10)+(2+3+4.............+11)(3+4........+12)+....................+(10+11+12...........+19)+2
  =(2*2+3*3+4*4+5*5........+10*10+11*9+12*8+.........19*1)+2
  =(2^2+3^2+4^2+5^2......+10^2+(10+1)(10-1)+............(10+9)(10-9))+2
  =(2^2+3^2+4^2+.......10^2+(10^2-1^2)+(10^2-2^2)+..................+10^2-9^2)
  =10(10^2)-1+2
  =1001
  
• 7 years agoHelpfull: Yes(2) No(0)
• A/c to f(x) f(101)=2,f(102)=3......f(199)=19,f(200)=2
  sum of f(100) to f(109)=1+2+3+4+5+6+7+8+9+10
  f(110) to f(119)=2+3+4+5+6+7+8+9+10+11=>1+2+3+4+5+6+7+8+9+10+10
  f(120) to f(129)=1+2+3+4+5+6+7+8+9+10+20
  . .
  . .
  f(190) to f(199)=1+2+3+4+5+6+7+8+9+10+90
  =>sum of f(100) to f(199)=10(55)+10(1+2+....+9)=10(55)+10(45)=1000
  sum of f(100) to f(200)=1000+2=1002
  sum of f(101) to f(200)=1002-1=1001
  then ans:- option 4:1001
• 7 years agoHelpfull: Yes(1) No(0)
• 10*sum of n natural numbers=10*n(n+1)/2
  =10*20*21/2
  =2100
• 7 years agoHelpfull: Yes(1) No(2)
• adding all the digits drom 101 to 200 so the answer is 1000
• 7 years agoHelpfull: Yes(0) No(7)