C programming - Conversion problems

Conversion Problems :-

1.      Convert each of the following binary numbers to octal, decimal, and hexadecimal formats.

®    (111011101)2

to octal: 111 011 101 = (735)8

to decimal: =(1x28) + (1x27) + (1x26) + (1x24) + (1x23) + (1x22)  + (1x20)

                        = 256 + 128 + 64 + 16 + 8 + 4 + 1

= (477)10

to hexadecimal: 0001 1101 1101 = (1DD)16

®    (10101010111)2

to octal: 010   101   010   111 = (2527)8

to decimal: =(1x210) + (1x28) + (1x26) + (1x24) + (1x22)

 + (1x21) + (1x20)

= 1024 + 256 + 64 + 16 + 4 + 2 + 1

= (1367)10

to hexadecimal: = 0101   0101   0111 =  (557)16

®    (111100000)2

to octal: = 111   100   000  = (740)8

to decimal: =(1x28) + (1x27) + (1x26) + (1x25)

= 256 + 128 + 64 + 32

= (480)10

to hexadecimal: = 0001   1110     0000  = (1E0)16

2.      Convert each of the following octal numbers to binary, decimal, and hexadecimal formats.

®    (3754)8

to binary:   = (11 111 101 100)2

to decimal:            =(3x83) + (7x82) + (5x81) + (4x80)

= 1536 + 448 + 40 + 4

= (2028)10

to hexadecimal:     = (0111 1110 1100)2

= (7EC)16

®    (7777)8

to binary: = (111 111 111 111)2

to decimal =(7x83) + (7x82) + (7x81) + (7x80)

= 3584 + 448 + 56 + 7

= (4095)10

to hexadecimal: = (1111 1111 1111)2

= (FFF)16

®    (247)8

to binary: = (10 100 111)2

to decimal: =(2x82) + (4x81) + (7x80)

= 128 + 32 + 7

= (167)10

to hexadecimal: = (1010 0111)2

= (A7)16

3.      Convert each of the following decimal numbers to binary, octal, and hexadecimal formats.

®    (3479)10

to binary: = 3479 ÷2 = 1739

rem = 1

1739 ÷2 = 869

rem = 1

869 ÷2 = 434

rem = 1

434 ÷2 = 217

rem = 0

217 ÷2 = 108

rem = 1

108 ÷2 = 54

rem = 0

54 ÷2 = 27

rem = 0

27 ÷2 = 13

rem = 1

13 ÷2 = 6

rem = 1

6 ÷2 = 3

rem = 0

3 ÷2 = 1

rem = 1

1 ÷2 = 0

rem = 1

reading bottom to top of remainders = (110110010111)2

to octal: = 3479 ÷8 = 434

rem = 7

434 ÷8 = 54

rem = 2

54 ÷8 = 6

rem = 6

6 ÷8 = 0

rem = 6

reading bottom to top of remainders = (6627)8

to hexadecimal: = 3479 ÷16 = 217

rem = 7

217 ÷16 = 13

rem = 9

13 ÷16 = 0

rem = 13 (D)

reading bottom to top of remainders = (D97)16

®    (642)10

to binary: = 642 ÷2 = 321

rem =0

321 ÷2 = 160

rem = 1

160 ÷2 = 80

rem = 0

80 ÷2 = 40

rem = 0

40 ÷2 = 20

rem = 0

20 ÷2 = 10

rem = 0

10 ÷2 = 5

rem = 0

5 ÷2 = 2

rem = 1

2 ÷2 = 1

rem = 0

1 ÷2 = 0

rem = 1

reading bottom to top of remainders = (1010000010)2

to octal: = 642 ÷8 = 80

rem = 2

80 ÷8 = 10

rem = 0

10 ÷8 = 1

rem = 2

1 ÷8 = 0

rem = 1

reading bottom to top of remainders = (1202)8

to hexadecimal: = 642 ÷16 = 40

rem = 2

40÷16 = 2

rem = 8

2 ÷16 = 0

rem = 2

reading bottom to top of remainders = (282)16

®    (555)10

to binary: = 555 ÷2 = 277

rem = 1

277 ÷2 = 138

rem = 1

138 ÷2 = 69

rem = 0

69 ÷2 = 34

rem = 1

34 ÷2 = 17

rem = 0

17 ÷2 = 8

rem = 1

8 ÷2 = 4

rem = 0

4 ÷2 = 2

rem = 0

2 ÷2 = 1

rem = 0

1 ÷2 = 0

rem = 1

reading bottom to top of remainders = (1000101011)2

to octal: = 555 ÷8 = 69

rem = 3

69 ÷8 = 8

rem = 5

8 ÷8 = 1

rem = 0

1 ÷8 = 0

rem =1

reading bottom to top of remainders = (1053)8

to hexadecimal: = 555 ÷16 = 34

rem = 11 (B)

34 ÷16 = 2

rem = 2

2 ÷16 = 0

rem = 2

reading bottom to top of remainders = (22B)16

4.      Convert each of the following hexadecimal numbers to binary, octal, and decimal formats.

®    (4FB2)16

to binary: (100 1111 1011 0010)2

to octal: (100 1111 1011 0010)2

= (47662)8

to decimal: = (4x163) + (15x162) + (11x161) + (2x160)

= (4x4096) + (15x256) + (11x16) + (2x1)

= 16384 + 3840 + 176 + 2

= (20402)10

®    (88BAE)16

to binary: (1000 1000 1011 1010 1110)2

to octal: (10 001 000 101 110 101 110)2

= (2105656)8

to decimal: = (8x164) + (8x163) + (11x162) + (10x161) + (14x160)

= (8x65536) + (8x4096) + (11x256) + (10x16) + (14x1)

= 16384 + 3840 +176 + 14

= (560046)10

®    (DC4)16

to binary: (1101 1100 0100)2

to octal: (110 111 000 100)2

= (6704)8

to decimal: = (13x162) + (12x161) + (4x160)

= (13x256) + (12x16) + (4x1)

= 3328 + 192 + 4

= (3524)10

Fractional numbers:

5.      Conversion from binary number system to decimal system

111.11₂            = 1 x 2² + 1 x 2¹+ 1 x 2⁰ + 1 x 2^-1 + 1 x 2^-2

                        =    4     +    2     +    1     +    1/2   +   ¼   =  7.75₁₀

Conversion from decimal number system to binary system

7.75₁₀= (?)₂

6.      Conversion of the integer part

7₁₀=( 111)₂

Conversion of the fractional part:

perform a repeated multiplication by 2 and extract the integer part of the result

            0.75 x 2 =1.50 Þextract 1

            0.5 x 2 = 1.0    Þ extract 1                                                    0.75₁₀ = 0.11₂

            0.0              Þ stop

             \  Combine the results from integer and fractional part, 7.75₁₀ = 111.11₂

7.      Convert (0.625)₁₀  to its binary form

Solution:          0.625 x 2 = 1.25          Þ extract 1

                        0.25 x 2 = 0.5 Þ extract 0

                        0.5 x 2 = 1.0    Þ extract 1

                        0.0       Þ stop

\ (0.625)₁₀ = (0.101)₂

8.      Convert (0.6)₁₀  to its binary form

0.6 x 2 = 1.2    Þ extract 1

0.2 x 2 = 0.4    Þ extract 0

0.4 x 2 = 0.8    Þ extract 0

0.8 x 2 = 1.6    Þ extract 1

0.6 x 2 =          Þ        ( same as 1^st one )       

\ (0.6)₁₀ = (0.1001 1001 1001 …)₂

9.      Convert (0.8125)₁₀  to its binary form

0.8125 x 2 = 1.625      Þ extract 1

 0.625 x 2 = 1.25         Þ extract 1

0.25 x 2 = 0.5 Þ extract 0

0.5 x 2 = 1.0    Þ extract 1

0.0       Þ stop

\ (0.8125)₁₀ = (0.1101)₂

 Credits-- Madhan