Decimal-to-binary by repeated division When converting a positive decimal integer to binary using repeated division by 2, the first remainder obtained corresponds to which bit of the final binary result?

Introduction / Context:
Repeated division by 2 is a standard manual algorithm to convert decimal integers to binary. Understanding which end of the binary number each remainder represents avoids reversed outputs.

Given Data / Assumptions:

• The method divides by 2 repeatedly and records the remainder at each step.
• Remainders are 0 or 1.
• The process stops when the quotient becomes 0.

Concept / Approach:
Every division by 2 extracts the least significant bit of the current quotient. Therefore, the first remainder corresponds to the least significant bit (LSB). Writing remainders from last to first reconstructs the binary number from MSB to LSB.

Step-by-Step Solution:

Verification / Alternative check:
Example: N=13. Remainders: 13 mod 2=1 (LSB), 6 mod 2=0, 3 mod 2=1, 1 mod 2=1 → bits (from first to last remainder) 1,0,1,1 ⇒ reversed 1101₂, which is correct.

Why Other Options Are Wrong:

• MSB choices suggest the first remainder is the MSB, which is incorrect.
• The conditional statements add confusion; the algorithm always yields LSB first.

Common Pitfalls:
Writing remainders in the order generated without reversing, which results in a mirrored binary number.

Final Answer:
LSB