z is the no in decimal system such that Z=260*1024+73*512+128*129+81+9. Let Y be the octal representation of Z. how many times the digit 3 be there in Y.
a) 2
b) 3
c) 4
d) 5

• ans is 2
  
  z= 260*2*8^3 + 73*8^3+ 129*2*8^2 + 81 + 9
  z=593*8^3 + 258*8^2 + 88 + 2
  dividing z by 8 we get remainder as 2 which forms Lsb of octal number and qoutient is
  q1 = 593*8^2 + 258*8 + 11
  dividing q1 by 8 we get remainder as 3 which forms second last digit of number and qoutient q2 as
  q2 = 593*8 + 258 + 1 = 593*8 + 256 +3
  dividing by 8 we get remainder as 3 which forms next digit of octal number and qoutient q3 as
  q3 = 593+32
  on solving further we see that there are only these two 3 that come as remainder...
  so we get two 3 in the octal number
  
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