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+63 votes

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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

Correct Answer: $C$

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

Correct Answer: $C$

+41 votes

We have, 3*4096 + 15*256 + 5*16 + 3

= (2+1)*2^{12 }+ (8+4+2+1) * 2^{8} + (4+1)*2^{4 }+ 2 + 1

= 2^{13} + 2^{12 }+ 2^{11 }+ 2^{10 }+ 2^{9 }+ 2^{8} + 2^{6} + 2^{4} + 2 + 1

= 1 1 1 1 1 1 1 1 1 1

= 10 1's So, **OPTION (C) .. **

+28 votes

$3 = (11)_2$

$3\times 4096 = 3\times (2^{12}) = (11)_2<< 12 = (11000000000000)_2$

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.

$3\times 4096 = 3\times (2^{12}) = (11)_2<< 12 = (11000000000000)_2$

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.

+4 votes

Answer is C.

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

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

0 votes

Powers of 16 go like: 1,16, 256, 1024, 65536, 1048576...

Interestingly enough, these are all powers of 2 as well.

As it happens, even the powers of 8 conform to this pattern.

Powers of 8 go like: 1, 8, 64, 512, 4096, 32768

That's because 16 is nothing but $2^4$; so $16^2$ = $(2^4)^2$ = $2^8$. Same can be derived for 8.

So, when you see 4096, don't immediately jump to $2^{12}$ — first check if it is a power of 16 or 8 as well. Because higher the base, easier is to solve such questions.

Coming to the question, given is hexadecimal format => 0x3F53 = Ten 1's in binary.

**Option C**

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