what are the total numbers which are < 1000 and the number neither be square nor cube....

a) 961 b)962 c)963 d)965

• for sqr 1 to 31 = 31 numbers
  for cube 8,27,125,216,343,512= 6 numbers
  therefore 999-37=962numbers are there
  
• 10 years agoHelpfull: Yes(47) No(10)
• square= 31
  cubes=9
  total 31+9=40
  now there are three such numbers which are square and cubes both 1, 64 and 729
  there total numbers which are either square or cube= 40-3= 37
  hence total numbers which are neither square nor cube= 999-37=962
• 10 years agoHelpfull: Yes(31) No(1)
• its 962.
  999-(31+6)=962
  
• 10 years agoHelpfull: Yes(6) No(1)
• total square no.- 31
  total quab no.- 9 (because 1000 is not include)
  but 1,64,729 has square and quab both so subtract 3
  31+9-3=37
  remains 999-37=962
• 10 years agoHelpfull: Yes(4) No(1)
• Square are from 1 to 31 = 31 no.s
  Cubes are from 2 to 9 = 7 no.s
  (1 is not taken because it is already taken in square field and 10's cube is 1000 and we have to
• 10 years agoHelpfull: Yes(1) No(2)
• @suchit
  9^3 = 27^2 = 729
  
• 10 years agoHelpfull: Yes(1) No(1)
• total sqr=30(31-1)
  cube=7(9-2)
  total=37
  ans=999-37=962
• 10 years agoHelpfull: Yes(1) No(0)
• a 961
  1 to 31 suare
  and
  cube is 8,27,125,216,343,512
• 10 years agoHelpfull: Yes(0) No(3)
• ans will be 961.
  explanation:
  last number whose sq
• 10 years agoHelpfull: Yes(0) No(2)
• @shivam. ya buddy u r right. thanks.
• 10 years agoHelpfull: Yes(0) No(0)
• total number of the number whose square is
• 9 years agoHelpfull: Yes(0) No(0)
• why 9^3=729 nt considered here....nly till 8^3 was considerd
  
• 9 years agoHelpfull: Yes(0) No(0)