8-bit representation question

Question

Using 8-bit representation compute the following sums
a) 111 + 45
b) 67 + 67

Answer 1

8 bit representation = 2⁷2⁶2⁵2⁴2³2²2¹2⁰

 

111 (decimal) = 2⁶ (=64) + 2⁵ (=32) + 2³ (=8) + 2² (=4) + 2¹ (=2) + 2⁰ (=1) = 0110 1111

 

45 (decimal) = 2⁵ (=32) + 2³ (=8) + 2² (=4) + 2⁰ (=1) = 0010 1101

 

Add:

 

  0110 1111    

+0010 1101   Remember 0+0= 0, 0+1=1, 1+0=1, 1+1=10 (you'll have to carry the one)

  ------------

  1001 1100

 

Check:

 

1011 1100 = 2⁷ (=128) + 2⁴ (=16) + 2³ (=8) + 2² (=4)  = 156

 

111 + 45 = 156

 

Try problem (b) on your own and let me know what you get (in the comments).  If you have questions, ask in the comments.

Answer 2

Step 1) Change number to 8-bit

Ex:

111 = 64+32+4+2+1 = 2⁶+2⁵+2²+2¹+2⁰

 

111 is 01101111

 

Step 2) Sum, note carry over method. No digit can be greater than 1

Example 001 + 001

 

 

    1

  001

+001

 010

 

a)       1 1    1 1 1 1            <--- this is carry over

(111)  0 1 1 0 1 1 1 1

(45)    0 0 1 0 1 1 0 1

          1 0 0 1 1 1 0 0

 

Check answer

2⁷+2⁴+2³+2² =156

 

b)

(67) 01000011

(67) 01000011

       10000110

 

This answer has been encoded using an altered Caesar Cipher.

(67) PJPPPPJJ

(67) PJPPPPJJ

       JPPPPJJP

 

Daniel, I challenge you to write a script/function to convert a number less than 256 into a string of zeros and ones representing the 8-bit format.

Answer 3

Using 8-bit representation compute the following sums
a) 111 + 45
b) 67 + 67

a)

111
  55 1
  27 1
  13 1
    6 1
    3 0
    1 1
    0 1

111_10 = 01101111_2

45
22 1
11 0
  5 1
  2 1
  1 0
  0 1

45_10 = 00101101_2

                 1101111
111_10 = 01101111_2
  45_10 = 00101101_2
–––––––––––––––––––
156_10 = 10011100_2

check:
      1 = 1
)2+0 = 2
)2+0 = 4
)2+1 = 9
)2+1 = 19
)2+1 = 39
)2+0 = 78
)2+0 = 156 √

b)

67
33 1
16 1
  8 0
  4 0
  2 0
  1 0
  0 1

67_10 = 01000011_2

  1             1        11
  67_10 = 01000011_2
  67_10 = 01000011_2
–––––––––––––––––––
134_10 = 10000110_2

check:

     1 = 1
)2+0 = 2
)2+0 = 4
)2+0 = 8
)2+0 = 16
)2+1 = 33
)2+1 = 67
)2+0 = 134 √

Notice that if 01000011_2 is left shifted by 1 bit you get 10000110_2; so a left shift of n bits is the same as multiplying by 2^n.

Answer 4

A) 111 = 1*2^6+1*2^5+1*2^3+1*2^2+1*2^1+1*2^0=1101110

45 = 1*2^5+1*2^3+1*2^2+1*2^0=101101

So the sum is as follows:

 

Strting with the smallest digit 0+1=1

Then 1+0=1

Then 1+1=0 carry one.

Then 1+1+1=1 carry one

Then 0+0+1=1

then 1+1=0 carry one.

Then 1+1=0 carry one

Thenfirst digit is one.

 

Sum is 1001101

B) 67 base ten = 1*2^6+1*2^1+1*2^0=1000011

1000011+1000011=10000110

Answer 5

' Octal representation

 

111 = ' 157              / Test '157 = 7 + 8 *5 + 8^2 = 111

 

  47  = '57              / 7 + 5 *8 = 47 

 158 =  '236          /  2* 64 + 3 *8 + 6 = 158

 

    ' 157 +' 57  = '236  

Binary:

     01101111 +00 101111 = 10011110

 

    67 =   '103         /  3 + 1*8^2 = 67

 

      103' + 103' = 20 6'    / 6 + 2 * 8^2 = 134

 

 Binary :

          1000011 + 1000011 = 10000110