Menghitung Bilangan Binner

1. NIM : 2014112007
Nama : Fahrul Ilhami
Jurusan : TI 1 / 2
Mata Kuliah : Arsitektur dan Organisasi Komputer
Konversikan !
1. (1024)10 = ( ?)2 = ( ?)8 = ( ?)16
2. (7FFC)16 = ( ?)2 = ( ?)10 = ( ?)8
3. (2472)8 = ( ?)10 = ( ?)2 = ( ?)16
4. (1111000111011)2 = ( ?)10 = ( ?)8 = ( ?)16
Jawaban
1. (1024)10 = ( ?)2 = ( ?)8 = ( ?)16
a. (1024)10 = ( ? )2
1024/2 = 512 0 LSB
512/2 = 256 0
256/2 = 128 0
128/2 = 64 0
64/2 = 32 0
32/2 = 16 0
16/2 = 8 0
8/2 = 4 0
4/2 = 2 0
2/2 = 1 0
MSB
Hasilnya adalah (10000000000)2
b. (1024)10 = ( ? )8
1024/8 = 128 0 LSD
128/8 = 16 0

2. 16/8 = 2 0
MSD Hasilnya adalah (2000)8
c. (1024)10 = ( ? )16
1024/16 = 64 0 LSD
64/16 = 4 0
MSD
Hasilnya adalah (400)16
2. (7FFC)16 = ( ? )2 = ( ? )10 = ( ? )8
a. (7FFC)16 = ( ?)2
= ( 7 = 0 1 1 1 ) + ( 15 = 1 1 1 1 ) + ( 15 = 1 1 1 1 ) + ( 12 = 1 1 0 0 )
= 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0
= (1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 )2
b. (7FFC)16 = ( ?)10
= ( 7 x 163 ) + ( 15 x 162 ) + ( 15 x 161 ) + ( 12 x 160 )
= ( 7 x 4096 ) + ( 15 x 256 ) + ( 15 x 16 ) + ( 12 x 1 )
= (28672) + (3840) + (240) + (12)
= ( 32764 )16
c. (7FFC)16 = ( ? )2
= = ( ? )8
2
= ( 22 x 21 x 20 ) + ( 22 x 21 x 20 ) + ( 22 x 21 x 20 ) +( 22 x 21 x 20 ) + (
22 x 0 x 0)
= ( 7 + 7 + 7 + 7 + 4 )
= (32)8
3. (2472)8 = ( ?)10 = ( ?)2 = ( ?)16
a. (2472)8 = ( ?)10
= ( 2 x 83 ) + ( 4 x 82) + ( 7 x 81 ) + ( 2 x 80 )
= ( 2 x 512 ) + ( 4 X 64 ) + ( 7 x 8 ) + ( 2 x 1 )
= ( 1024 + 256 + 56 + 2
= ( 1338 )10
b. (2472)8 = ( ?)2
= ( 2 = 101 ) + ( 4 = 100 ) + ( 7 = 111 ) + ( 2 = 101 )
= ( 101 100 111 101 )2
c. (2472)8 = ( ?)16
= ( 101 100 111 101 )2 = ( ? )16
= 101 0011 1010
5 3 10
= ( 53A )16
111 111 111 111 100

3. 4. (1111000111011)2 = ( ? )10 = ( ? )8 = ( ? )16
a. (1111000111011)2 = ( ? )10
= ( 1 x 212 ) + ( 1 x 211 ) + ( 1 x 210 ) + ( 1 x 29 ) + ( 0 x 28 ) +
( 0 x 27 ) + ( 0 x 26 ) + ( 0 x 25 ) + ( 1 x 24 ) + ( 1 x 23 ) +
( 0 x 22 ) + ( 0 x 21 ) + ( 0 x 20 )
= ( 1 x 4096 ) + ( 1 x 2048 ) + ( 1 x 1024 ) + ( 1 x 512 ) +
( 0 x 256 ) + ( 0 x 128 ) + ( 0 x 64 ) + ( 1 x 32 ) + ( 1 x 16 )
+ ( 1 x 8 ) + ( 0 x 4 ) + ( 1 x 2 ) + ( 1 x 1 )
= ( 4096 ) + ( 2048 ) + ( 1024 ) + ( 512 ) + ( 0 ) + ( 0 ) + ( 0
) + ( 32 ) + ( 16 ) + ( 8 ) + ( 0 ) + ( 2 ) + ( 1 )
= (7739)10
b. (1111000111011)2 = ( ?)8
= ( 1 111 000 111 011 )2
= 1 7 0 7 3
= (1 7 0 7 3)8
c. (1111000111011)2 = ( ?)16
= ( 1 1110 0011 1011 )
= 1 14 3 11
= (1 E 3 B)16