how many no. are less then 1000 which are neither a square nor cube of any no.

• total no. less than 1000 are 999. In these no.s 31 no.s will be perfect square because 31^2=961 and 9 no.s will be perfec cube from 1^3 to 9^3 . so total no. which r perfect square nd perfect cube r 40 bt in these no. some nos r common which r perfect square nd perfect cube both 1,64,729 so total no. 40-3=37 so total no. which r neither perfect square nor perfect cube iz 999-37=962.
  so ans will be 962.
• 11 years agoHelpfull: Yes(55) No(0)
• Rupesh bhai KLPD ho gya.. u dint count 1 and I counted 1 twice... both our answers are wrong. Answer will be 962. :P :P
• 11 years agoHelpfull: Yes(8) No(1)
• ans:963
  in case of square:
  sequence like this:2^2,3^2..........................31^2 last term is 31^2 because 31^2=961 and if we calculate 32^2=1024 it exceed from 1000,so we can only calculate at 31^2,so here total no=30.
  now,in case of cube sequence like this:
  8,27,64,125,216,343,512,729.
  but in case of square 64 is a square of 8 and 729 is square of 27 and it is also cube of 4 and 9 respectively.so here total no is 6.
  so total no=999-(30+6)=963
• 11 years agoHelpfull: Yes(5) No(12)
• answer will be (999-38)=961
• 11 years agoHelpfull: Yes(2) No(7)
• 962
  
  1-1000 => 10 cube
  1-1000 => 31 sq
  common r 1,64, 729 ... so counted only 1ce....
  => 41-3=38 ... => 962 r not either of dem
• 11 years agoHelpfull: Yes(0) No(0)
• total no. less than 1000= 999.
  perfect squares= n^2..(1,4,9,16,25,36,49..upto 961) where n=1 to 31.so 31 numbers
  perfect cubes=n^3..(8,27,64,125,216,343,512,729) where n= 2,3,4,5,6,7,8,9.so 8 nos.
  now the nos which are both perfect squares and cubes..n^6..(64,729) where n=2,3.so 2 nos.
  
  total nos=999-(31+8)-2=962.
• 11 years agoHelpfull: Yes(0) No(0)
• its 961..
  because 64 is the square of 8 and cube of 4...
  so it has to be counted once..
  therefore..999-38=961
• 11 years agoHelpfull: Yes(0) No(0)