Z=260*1024+73*512+128*129+81+9 (z in decimal system).
now y be the octal representation of z in number system.How many time will the digit 3 be in y
options = >2/3/4/5

• Z=260*1024+73*512+128*129+81+9
  
  convert into power of 8
  260*1024 = 2*260*8^3= 65*8^4 = 8^4(8^2+1)= 8^6+8^4
  
  73*512 = 8^3*(72+1) = 8^3[8*(8+1)+1] = 8^3(8^2+8^1+1)= 8^5+8^4+8^3
  
  128*129 = 8^2*258 = 8^2(8*32+2)= 8^2(4*8^2+2)= 4*8^4+2*8^2
  
  81= 8^2+16+1 = 8^2+2*8^1+8^0
  
  9= 8^1+8^0
  
  now Z can be written as
  
  z = 8^6 + 8^5 + 6*(8^4) + 8^3 + 3*(8^2) + 3*(8^1) + 2*(8^0)
  
  z= octal(1161332)8
  
  in octal representation of Z = 1161332, no, of 3's are 2
  
  ans: 2
• 10 years agoHelpfull: Yes(35) No(0)
• z can be written in powers of 8(as octalconversion is reqd.)
  => 8^6+8^5+6*8^4+8^3+3*8^2+3*8^1+2*8^0
  so no of 3s are 2
  Ans: 2
• 10 years agoHelpfull: Yes(5) No(5)
• 260*1024 = 2*260*8^3= 65*8^4 = 8^4(8^2+1)= 8^6+8^4
  
  73*512 = 8^3*(72+1) = 8^3[8*(8+1)+1] = 8^3(8^2+8^1+1)= 8^5+8^4+8^3
  
  128*129 = 8^2*258 = 8^2(8*32+2)= 8^2(4*8^2+2)= 4*8^4+2*8^2
  
  81= 8^2+16+1 = 8^2+2*8^1+8^0
  
  9= 8^1+8^0
  
  now Z can be written as
  
  z = 8^6 + 8^5 + 6*(8^4) + 8^3 + 3*(8^2) + 3*(8^1) + 2*(8^0)
  
  z= octal(1161332)8
  
  in octal representation of Z = 1161332, no, of 3's are 2
  
  ans will be 2
• 10 years agoHelpfull: Yes(4) No(0)
• 260*1024 = 2*260*8^3= 65*8^4 = 8^4(8^2+1)= 8^6+8^4
  
  73*512 = 8^3*(72+1) = 8^3[8*(8+1)+1] = 8^3(8^2+8^1+1)= 8^5+8^4+8^3
  
  128*129 = 8^2*258 = 8^2(8*32+2)= 8^2(4*8^2+2)= 4*8^4+2*8^2
  
  81= 8^2+16+1 = 8^2+2*8^1+8^0
  
  9= 8^1+8^0
  
  now Z can be written as
  
  z = 8^6 + 8^5 + 6*(8^4) + 8^3 + 3*(8^2) + 3*(8^1) + 2*(8^0)
  
  z= octal(1161332)8
  
  in octal representation of Z = 1161332, no, of 3's are 2
  
• 10 years agoHelpfull: Yes(1) No(0)