Z is a number in decimal system such that Z=260*1024+73*512+128*129+81+9. Let Y be the octal representation of 'Z' (that is representation of 'z' in a number system to the base 8 is 'Y'). How many times will the digit 3 be there in 'Y'

• 260*1024+73*512+129*128+90->2
  260*128+73*64+129*16+11->3
  260*16+73*8+259->32
  625->1
  78->6
  9->1
  1->1
  
  reading from down to top 1161332.
  so there are two 3's.
• 11 years agoHelpfull: Yes(24) No(4)
• @Taashi how 2,3,32 are forming please explain
  
  
  
  
  
  
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• @ganesh when we dive by 8 the remaining sum is 2,3,3 its not 32 dont confuse
• 9 years agoHelpfull: Yes(0) No(0)
• Given,
  Z=260*1024+73*512+128*129+81+9(which is in decimal form)
  
  Convert each number in to octal form by placing remainders......
  Let's take for (260)base to the 10 = (404)base to the 8
  Similarily 1024= 2000
  ......
  
  Finally octal form of z is
  Z=404*2000+111*1000+200*201+121+11
  
  ---111=3*37 && 201=3*67
  
  Z=404*2000+(3*37)*1000+200*(3*67)+121+11
  
  How many times the digit 3 is repeated.......
  2 times
• 8 years agoHelpfull: Yes(0) No(1)