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The number of $1$'s in the binary representation of $(3*4096 + 15*256 + 5*16 + 3)$ are:

1. $8$
2. $9$
3. $10$
4. $12$
edited | 5.8k views

I suggest following approach , here we can clearly see that numbers are getting multiplied by powers of 16. So this is nothing but Hexadecimal number in disguise.

$(3\times 4096 + 15 \times 256 + 5 \times 16 + 3) = (3F53)_{16} = (0011111101010011)_2$ which has total $2 + 4 + 2 + 2 = 10$ 1's
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awesome

We have,  3*4096 + 15*256 + 5*16 + 3

= (2+1)*212     + (8+4+2+1) * 28          +  (4+1)*24  + 2 + 1

= 213 + 212      + 211  + 210 + 2+ 28   +   26 + 24   + 2 + 1

= 1        1            1       1        1      1          1      1      1     1

= 10  1's     So, OPTION (C) ..

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Thanks @Himanshu it is best approach.
$3 = (11)_2$
$3\times 4096 = 3\times (2^{12}) = (11)_2<< 12 = (11000000000000)_2$

Similarly, $15 \times 256 = (1111)_2 << 8 = (111100000000)_2$ and $5 \times 16 = (101)_2 << 4 = (1010000)_2$

So, $3\times 4096 + 15 \times 256 + 5\times 16 + 3 = (11111101010011)_2$

Number of 1's = 10.
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best approach..

All 4096,256,16 needs only 1 one to be represented in binary

3 - requires 2   1's

15 - requires 4 1's

5 - requires 2 1's

3 - requires 2 1's

so adding all those we get 2+4+2+2= 10
+1
10 is correct but the approach works only if the numbers doesn't have a 1 in the same place. You have to take care of that.

((3*4096) +(15*256) + (5*16) + 3 ) = 16211

= (11111101010011)2

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