N is a natural number of at least 5 digits and its leftmost digit is 6. When this 6 is removed from N, the number thus obtained is found to be 1/25 times of N. What is the sum of the digits of N?

• 13
  
  number can be 62500, 625000, .....
• 12 years agoHelpfull: Yes(4) No(0)
• Let x be the no. & no. of digits be n*10000 (atleast 5 digits)
  
  x-60000*(n) = x/25 hence x=62500*(n) where n=1,10,100,100 etc
  
  
• 12 years agoHelpfull: Yes(1) No(6)
• let the number be 6_ _ _.....
  
  this can be written as 6*(10^n) + _ _ _...
  
  where n >3 for the number to be atleast 5 digit........(1)
  
  let _ _ _... be m
  
  so the number is 6*(10^n) + m............(2)
  
  Now, (1/25)*(6*(10^n) + m)=m
  
  -->10^n=4*m
  
  from eqn 1 we put n=4 first and get m=2500
  
  hence substituting n,m in eqn 2 we get number as 62500
  
  so the sum of the digits =13(ANS)
  
• 8 years agoHelpfull: Yes(0) No(0)
• 10^4*6+10^3*x4+10^2*x3+10 x2+x1=N
  now 10^3*x4+10^2*x3+10 x2+x1=N/25
  10^4*6+N/25=N
  N=625000, 6+2+5=13
• 6 years agoHelpfull: Yes(0) No(0)