ISRO CSE 2007 | Question: 18

Question

The number of digit 1 present in the binary representation of

$3 \times 512 + 7 \times 64 + 5 \times 8 + 3$ is

• A. 8
• B. 9
• C. 10
• D. 12

Answer 1

Option B

3*512 + 7*64 +5*8+3

=(2+1)2⁹ + (2²+2+1)2⁶ +(2²+1)2³+(2+1)

=2¹⁰+2⁹+2⁸+2⁷+2⁶+2⁵+2³+2¹+2⁰

=11111101011

Therefore Total Number of 1's =9

Answer 2

Option B

The octal representation of the given number is 3753 (the given number is 3*8^3+7*8^2+5*8+3.so it is easy to represent in octal)

then octal to binary conversion.

011 111 101 011

Answer 3

3*512+7*64+5*8+3

As we can see that 

512=2^9 have one 1

64=2^6 have one 1

8=2^3 have one 1

Hence,they remains as same in no of 1's and when any other no is multiplied by themselves.

So, no of 1's is only the count of the no of 1's in other multiplicant. 

3=011    # of 1's=2

7=111    # of 1's=3

5=101    # of 1's=2

3=011    # of 1's=2

          ---------------------

    Total  # of 1's=9

         

Comments

superb solution

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Best explanation

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wait a min,

I get everything except that last line. How you get the no of 1s are 9?

Lets say a = 3 = 11 and b = 2 = 10. a + b = 100. So the number of 1s are not added...

Answer 4

9 ones

Option B is correct

Answer 5

(3 x 512) + (7 x 64) + (5 x 8) + 3

= (3 x 8³ ) + (7 x 8² )  + (5 x 8¹) + (3 x 8⁰)

It is the decimal representation of an octal number 3753 which is equivalent to " 011 111  101 011 " in binary which contains 9 1's..

So, answer is (b)