GATE CSE 1995 | Question: 2.12, ISRO2015-9

Question

The number of $1$'s in the binary representation of $(3\ast4096 + 15\ast256 + 5\ast16 + 3)$ are:

• A. $8$
• B. $9$
• C. $10$
• D. $12$

Comments

$4096=2^{12}$ and multiplying a number in binary with $2^x$ means that we are shifting that number to the left by $x$ bits.

So, the number $4096 . 3$ will be $(11)_2$ shifted by $12$ bits to the left which will become $11000000000000$.

Same we can do with other numbers as well and will find that there is no overlapping $1's$. So total $1's = 1's$ in $3,15,5,3$.

So the answer is $10$.

Result =

$11000000000000$ $+$

     $111100000000$ $+$

                 $1010000$ $+$

                             $11$

$=11111101010011$

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But what if the question is like 15*256 + 5*16 – 3?

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they are in hexadecimal notation write 3,15, 5,3  into 4 digits each

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@Nischal Mehta we dont include negative numbers in the notation

Answer 1

I suggest the 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.

Comments

awesome

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good observation

 

Answer 2

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 + 2⁹ + 2⁸   +   2⁶ + 2⁴   + 2 + 1

              = 1        1            1       1        1      1          1      1      1     1

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

Comments

Thanks @Himanshu it is best approach.

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Best Approach

Answer 3

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

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.

Comments

best approach..

Answer 4

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

Comments

10 is correct but the approach works only if no two numbers have a 1 in the same bit position. You have to take care of that.

Answer 5

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

= (11111101010011)2

Answer 6

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.

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

Option C

Comments

In place of 1024, 4096 will be there.