Question

question 2 of 10
how is the decimal number 107 expressed in binary code?

a. 1001100

b. 1101011

c. 1010110

d. 1100100

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Explanation:

Step1: Divide 107 by 2

$107 \div 2 = 53$ with a remainder of $1$ (least significant bit)

Step2: Divide 53 by 2

$53 \div 2 = 26$ with a remainder of $1$

Step3: Divide 26 by 2

$26 \div 2 = 13$ with a remainder of $0$

Step4: Divide 13 by 2

$13 \div 2 = 6$ with a remainder of $1$

Step5: Divide 6 by 2

$6 \div 2 = 3$ with a remainder of $0$

Step6: Divide 3 by 2

$3 \div 2 = 1$ with a remainder of $1$

Step7: Divide 1 by 2

$1 \div 2 = 0$ with a remainder of $1$ (most significant bit)

Step8: Read remainders from last to first

Combining the remainders from step 7 to step 1: $1101011$

We can also verify by converting the binary options back to decimal:

• Option A: $1\times2^6 + 0\times2^5 + 0\times2^4 + 1\times2^3 + 1\times2^2 + 0\times2^1 + 0\times2^0 = 64 + 8 + 4 = 76

eq 107$

• Option B: $1\times2^6 + 1\times2^5 + 0\times2^4 + 1\times2^3 + 0\times2^2 + 1\times2^1 + 1\times2^0 = 64 + 32 + 8 + 2 + 1 = 107$
• Option C: $1\times2^6 + 0\times2^5 + 1\times2^4 + 0\times2^3 + 1\times2^2 + 1\times2^1 + 0\times2^0 = 64 + 16 + 4 + 2 = 86

eq 107$

• Option D: $1\times2^6 + 1\times2^5 + 0\times2^4 + 0\times2^3 + 1\times2^2 + 0\times2^1 + 0\times2^0 = 64 + 32 + 4 = 100

eq 107$

Answer:

B. 1101011