How many 1's are there in the binary form of  8*1024 + 3*64 + 3

4

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dhanasekaran

  • Dec 6th, 2005
 

the binary eqhivalent of 8387 is as follows10000011000011

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Atul Dongare

  • Apr 17th, 2006
 

I think answer= 4 1's

it is because when we convert 8387 into binary then ans will be 10100011

so the no. of ones are 4

ans is 4

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Parag

  • Apr 19th, 2006
 

Due to weight of each bit we have;8=1 one in binary form1024=1 one ie for 2 raise to 10 th position 3=2 ones and there are 2 threes in expression therefore 4 ones64=1 one means total 7 ones.Why to convert to binary unnneccesarily?

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8*1024 + 64*3 + 3 = 8387
 8387/2 then remainder = 1
 4193/2 then remainder = 1
 2096/2 then remainder = 0
 1048/2 then remainder = 0
 524/2 then remainder = 0
 262/2 then remainder = 0
 131/2 then remainder = 1
 65/2 then remainder = 1
 32/2 then remainder = 0
16/2 then remainder = 0
 8/2 then remainder = 0
4/2 then remainder = 0
2/2 then remainder = 0
1/2=0 then remainder = 1

8387= 10000011000011

so overall we have 5 1's
Thank you

ANS is 5.
8387/2=4193              Remainder 1
4193/2=2096                 1
2096/2=1048                 0
1048/2=524                   0
524/2=262                     0
262/2=131                     0
131/2=65                       1
65/2=32                         0
32/2=16                         0
16/2=8                           0
8/2=4                             0
4/2=2                             0
2/2=1                             0
1                                    1     
so the binary code is 10000011000011
So Total no. of 1's =5

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ashok815

  • Dec 22nd, 2009
 

8*1024 + 3*64 + 3= (2^3*2^10) + (2^1+2^0)*2^6 + (2^1+2^0)
                              
                              = 2^13 + 2^7 +2^6 + 2^1 + 2^0


 Hence 5

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