Problemas sumas y restas de binarios

Dávila Pérez Tania Paulina Grupo: 5101

Realizar las siguientes conversiones usando tablas de equivalencias binario-octal, binario-hexadecimal a) 1001000111010100100010.0101(2) a octal
21 18 13 1210 8

= 2384282.20(8)
5 1

1x 2 + 1x 2 + 1x 2 + 1x 2 + 1x 2 + 1x 2 + 1x 2 + 1x 2 2097152+262144+16384+8192+4096+1024+256+32+2= 2384282
2 4

1x 2 + 1x 2 4+16= 20 b) 4EC7.B5(8) a binario414127.115= 1111010101110.000111
_ 207063 2 | 414127 014 012 07 1

103531 2 |207063 007 10 06 03 1 6475 2 |12941 09 14 11 1

51765 2 |103531 03 15 13 11 1 3237 2 |6475 04 07 15 1 1618 2 |3237 12 03 171

25882 2 |51765 11 17 16 05 1 809 2|1618 018 0

12941 2 |25882 05 18 08 02 0

404 2 |809 009 1

202 2 |404 004 0

101 2|202 002 0

50 2|101 01 1

15 2 |50 10 0

7 2|15 1

3 2 |7 11 2| 3 1

0 2|1 1

x

.115 2

x

.230 2

.460 x 2

x

.920 2

.840 x 2

.680 x 2

0.230 c)

0.460

0.920

1.840

1.680

1.360

475320.47(8) a hexadecimal
5 4 3 21

= 101560.751
0

4x 8 + 7x 8 + 5x 8 + 3x 8 + 2x 8 + 0 x 8 131072 + 28672 + 2560 + 192 + 16 + 0= 162512 .47 x 16 282 47__ 7.32 .32 x16 192 32_ 5.12 .12 x 16 72 12__ 1.92

d) 32FE685.9C(16) aoctal
6 5 4

= 66830565.817
3 2 1 0

3x 16 + 2x 16 + 15x 16 + 14x 16 + 6x 16 + 8x 16 + 5x 16 50331648 + 2097152 + 983040 + 57344 + 1536 + 128 + 5= 53470853

.9C x 8 8.24

.24 x 8 1.92

.92x 8 7.36

1.3 Convertir usando el método general a) 730568.23(9) a base 14
5 4 3 2 1 0

7x9 +3x9 +0x9 +5x9 + 6x9 +8x9 413343 + 19683 + 405 + 54 + 8= 433493 .23 x 14 92 23 3.12 .12 x 14 48 121.68 .68 x14 272 68_ 9.5

b) 6G5A.23(20) a binario 308255 2 |616510 016 05 11 10 0

= 01100110000101101001 _77068 2 |154127 14 012 07 1 38534 2 |77068 17 10 06 08 0

154127 2 |308255 10 08 02 0515 1 9633 2 |19267 12 06 07 1

19267 2 |38534 18 05 13 14 0

4816 2 |9633 16 03 13 1

2408 2 |4816 08 016 0

1204 2 |2408 04 008 0

_602 2 | 1204 004 0

301 2 |602 002 0

150 2 |301 10...