Question

Q.3. Convert ( 47 )10 to binary using repeated division method.​

Answer 1

Answer:

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Answer 2

Let’s illustrate that with the initial value 188

• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 0
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 0
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 1
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 1
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 1
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 1
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 0
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1

Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1Our first number is finally down to zero (0 R 0), so we read up the remainder column to see that the binary number is 10111100

Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1Our first number is finally down to zero (0 R 0), so we read up the remainder column to see that the binary number is 10111100Let’s look at the same thing in binary:

• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1Our first number is finally down to zero (0 R 0), so we read up the remainder column to see that the binary number is 10111100Let’s look at the same thing in binary:10111100 ÷ 2 = 1011110 R 0
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1Our first number is finally down to zero (0 R 0), so we read up the remainder column to see that the binary number is 10111100Let’s look at the same thing in binary:10111100 ÷ 2 = 1011110 R 01011110 ÷ 2 = 101111 R 0
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1Our first number is finally down to zero (0 R 0), so we read up the remainder column to see that the binary number is 10111100Let’s look at the same thing in binary:10111100 ÷ 2 = 1011110 R 01011110 ÷ 2 = 101111 R 0101111 ÷ 2 = 10111 R 1
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1Our first number is finally down to zero (0 R 0), so we read up the remainder column to see that the binary number is 10111100Let’s look at the same thing in binary:10111100 ÷ 2 = 1011110 R 01011110 ÷ 2 = 101111 R 0101111 ÷ 2 = 10111 R 110111 ÷ 2 = 1011 R 1
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1Our first number is finally down to zero (0 R 0), so we read up the remainder column to see that the binary number is 10111100Let’s look at the same thing in binary:10111100 ÷ 2 = 1011110 R 01011110 ÷ 2 = 101111 R 0101111 ÷ 2 = 10111 R 110111 ÷ 2 = 1011 R 11011 ÷ 2 = 101 R 1
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1Our first number is finally down to zero (0 R 0), so we read up the remainder column to see that the binary number is 10111100Let’s look at the same thing in binary:10111100 ÷ 2 = 1011110 R 01011110 ÷ 2 = 101111 R 0101111 ÷ 2 = 10111 R 110111 ÷ 2 = 1011 R 11011 ÷ 2 = 101 R 1101 ÷ 2 = 10 R 1
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1Our first number is finally down to zero (0 R 0), so we read up the remainder column to see that the binary number is 10111100Let’s look at the same thing in binary:10111100 ÷ 2 = 1011110 R 01011110 ÷ 2 = 101111 R 0101111 ÷ 2 = 10111 R 110111 ÷ 2 = 1011 R 11011 ÷ 2 = 101 R 1101 ÷ 2 = 10 R 110 ÷ 2 = 1 R 0
• Let’s illustrate that with the initial value 188188 ÷ 2 = 94 R 094 ÷ 2 = 47 R 047 ÷ 2 = 23 R 123 ÷ 2 = 11 R 111 ÷ 2 = 5 R 15 ÷ 2 = 2 R 12 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1Our first number is finally down to zero (0 R 0), so we read up the remainder column to see that the binary number is 10111100Let’s look at the same thing in binary:10111100 ÷ 2 = 1011110 R 01011110 ÷ 2 = 101111 R 0101111 ÷ 2 = 10111 R 110111 ÷ 2 = 1011 R 11011 ÷ 2 = 101 R 1101 ÷ 2 = 10 R 110 ÷ 2 = 1 R 01 ÷ 2 = 0 R 1