Number System: An Introduction

(267)₈

Step 1: Group the binary digits in sets of 3 from the right: 10 | 110 | 111. Step 2: Add a leading zero to the leftmost group to make it 3 bits: 010 | 110 | 111. Step 3: Convert each 3-bit group to its octal digit: 010=2, 110=6, 111=7. Step 4: Write the octal number: (267)₈. Option B (276) swaps the middle and last digits incorrectly. Option C (237) misreads the first group. Always group from right to left and add leading zeros if needed.

(101010110)₂

Step 1: Convert each octal digit to its 3-bit binary equivalent using the standard table. Step 2: 5 in octal = 101 in binary (since 4+1=5, so bits are 1,0,1). 2 in octal = 010 in binary. 6 in octal = 110 in binary (since 4+2=6). Step 3: Combine: 101 | 010 | 110 = (101010110)₂. Step 4: Leading zeros at the start are dropped: result is (101010110)₂. Option B has the wrong 3-bit code for 5. Option C misrepresents the value of 5 in binary. Option D changes the last group incorrectly.

(ABCD)₁₆

Step 1: Divide 43981 by 16. 43981÷16=2748 R13(D). Step 2: 2748÷16=171 R12(C). Step 3: 171÷16=10 R11(B). Step 4: 10÷16=0 R10(A). Step 5: Read remainders from bottom to top: A, B, C, D → (ABCD)₁₆. Remember: 10=A, 11=B, 12=C, 13=D in hex. This is a famous hex number! Option B swaps C and D. Option C completely rearranges the digits. Option D mixes up B and C.

(3882)₁₀

Step 1: Identify the hex digits: F=15, 2=2, A=10. Step 2: Assign positional powers of 16: F×16², 2×16¹, A×16⁰. Step 3: Calculate: 16²=256, 16¹=16, 16⁰=1. Step 4: Multiply: 15×256=3840, 2×16=32, 10×1=10. Step 5: Add: 3840+32+10 = 3882. So (F2A)₁₆ = (3882)₁₀. Option B (3928) uses the wrong value for A. Option C (3842) makes an error in 15×256. Option D uses wrong positional values.