Question

The binary representation of the decimal number 125 is

• A. 1110001
• B. 1010101
• C. 1111101
• D. 1111001

Hint:

When converting from decimal to binary, repeatedly divide by 2 and use the remainders to form the binary number.

Answer 1

The Correct Option is C

Step 1: Conversion from decimal to binary.
To convert 125 to binary, we divide the number by 2 and record the remainders: \[ 125 \div 2 = 62 \quad \text{remainder} \, 1 \] \[ 62 \div 2 = 31 \quad \text{remainder} \, 0 \] \[ 31 \div 2 = 15 \quad \text{remainder} \, 1 \] \[ 15 \div 2 = 7 \quad \text{remainder} \, 1 \] \[ 7 \div 2 = 3 \quad \text{remainder} \, 1 \] \[ 3 \div 2 = 1 \quad \text{remainder} \, 1 \] \[ 1 \div 2 = 0 \quad \text{remainder} \, 1 \] Reading the remainders from bottom to top, we get \( 1111101_2 \).
Step 2: Conclusion.
Therefore, the binary representation of 125 is 1111101.

Answer 2

The question asks for the binary representation of the decimal number 125. Rather than dividing repeatedly by 2, we can build the binary form top-down by greedily subtracting the largest available power of 2, and then check each option by converting it back to decimal.

Starting from 125: the largest power of 2 not exceeding 125 is \( 2^6=64 \), leaving \( 125-64=61 \). The largest power of 2 not exceeding 61 is \( 2^5=32 \), leaving \( 61-32=29 \). The largest power not exceeding 29 is \( 2^4=16 \), leaving \( 29-16=13 \). The largest power not exceeding 13 is \( 2^3=8 \), leaving \( 13-8=5 \). The largest power not exceeding 5 is \( 2^2=4 \), leaving \( 5-4=1 \). Then \( 2^1=2 \) does not fit into 1, so that bit is 0, and finally \( 2^0=1 \) fits exactly, leaving 0. So the bits used are \( 2^6,2^5,2^4,2^3,2^2,2^0 \), giving \( 1111101 \).

1. 1110001: Converting back, \( 64+32+16+0+0+0+1=113 \), not 125.
2. 1010101: Converting back, \( 64+0+16+0+4+0+1=85 \), not 125.
3. 1111101: Converting back, \( 64+32+16+8+4+0+1=125 \), matching exactly, consistent with the bits derived above.
4. 1111001: Converting back, \( 64+32+16+8+0+0+1=121 \), not 125.

Only the third option reconstructs to exactly 125 when its bits are added up, matching the greedy power-of-2 breakdown derived directly from 125.

Therefore, the correct answer is 1111101.