ME Test Series

Question

The number of 1’s present in binary representation of 3 * 4096 – 4 + 6 * 512 is ______

Comments

I am elaborating the explanation given by Made Easy:

$3*4096-4+6*512$

$=3*(2^{12}) - 4 +6*(2^9)$

$=3*(2^3)^4 - 4 + 6*(2^3)^3$

$=3*8^4+6*8^3-4*8^0$

$=(36000)8 - (00004)_8$

Now, $-(0004)_8$ = 8's complement of $(00004)_8$

8's complement of $(00004)_8$ = 7's complement of $(00004)_8$ + 1

7's complement of $(00004)_8$ = $(77777)_8 - (00004)_8 = (77773)_8$

8's complement of $(00004)_8$ = $(77773)_8+1 = (77774)_8$

So,

$=(36000)_8 - (00004)_8 = (36000)_8 + (77774)_8$

3 6 0 0 0

7 7 7 7 4

__________

 3  5 7 7  4

How?

the last 3 bits addition is okay.

6+7=13 = 1*8^1 + 5*8^0 so $(15)_8$, 5 is the sum and 1 is the carry here.

1+3+7=11= 1*8^1+3*8^3 = $(13)_8$ where 3 is the sum 1 is the carry. Discard the carry.

So, $(35774)_8$

Now expanding it in binary, 011 101 111 111 100 , we have 11 1's.

Reason to discard the carry https://stackoverflow.com/questions/22049866/why-should-i-discard-carry-out-in-adding-binary-numbers

Correct me if I am wrong

Answer 1

Given $3*4096-4+6*512$

Rewrite it as $3*8^{4}+6*8^{3}-4*8^{0}$

$(36000){_{8}}-(4){_{8}} $

$=(35774){_{8}}$

Convert it into binary

$=(011 \ 101\ 111\ 111\ 100){_{2}}$

Number of 1's is 11