if no of digit in X^100 is 31.then find the number of digit in
X^1000.

• ans:301
  if we have y digits in x^100, that means that
  10^(y-1) = x^100 [we put 10 to the (y-1) power because 10^1 has]
  [2 digits, 10^2 has 3 digits, so 10^(y-1) has ]
  [y digits ]
  in question we have 31 digits
  10^(31-1)=x^100
  applying log
  30=100logx ------>equ 1
  
  10^(y-1)=x^1000
  y-1=1000 logx
  y-1=10*100logx
  y-1=10*30
  y=301
  
  
  
  
• 11 years agoHelpfull: Yes(38) No(2)
• X^100 has 31 digits
  no. of digits in a no. can be found by taking its log
  i.e., log(X^100) = 31
  => 100 logX=31
  multiplying both sides by ten
  1000logX=310
  =>
  log(X^1000)=310
  no. of digits = 310
• 11 years agoHelpfull: Yes(30) No(32)
• Answer is 302 digits.
  
  10^(d−1) has d digits.
  So for any number n, the number of digits of n is given by solving 10^(d−1)=n, or
  
  d=1+⌊log10(n)⌋
  
  put x=2. It is satisfying the given condition, that is,
  1+100*⌊log10(2)⌋ = 1+100*⌊0.301⌋ = 1+⌊30.1⌋ = 1 + 30 = 31.
  
  So
  d = 1+1000*⌊log10(2)⌋ = = 1+1000*⌊0.301⌋ = 1+⌊301.something⌋ = 1 + 301 = 302
• 11 years agoHelpfull: Yes(10) No(4)
• solution by aruna is correct... 10^(y-1)=x^100 is a formulae...where y will stand for number of digits
• 11 years agoHelpfull: Yes(5) No(0)
• x^100=31 i.e (x^10)^10=31
  x^1000=(((x^10)^10)^10)=(31^10)
• 11 years agoHelpfull: Yes(4) No(12)
• x^100 have 31 digits => x = 2 only as 3^100 have more than 31. but for x=2 no of digit will be 30
  so, x^1000 = 2^1000 = y
  => logy = 1000*log2=1000*0.301
  => logy = 301
  so y has 301+1 = 302 digits.
• 10 years agoHelpfull: Yes(1) No(0)
• 101 has 2 digits = 10
  102 has 3 digits = 100
  103 has 4 digits = 1000
  1030 has 31 digits
  
  therefore, 1030=x100
  30=100logx
  multiplying both sides by 10
  30⋅10=10⋅100logx
  300=1000logx=300+1=301
• 6 years agoHelpfull: Yes(0) No(0)