[https://www.youtube.com/watch?v=eaPAUYoM-r0:embed:cite]
The program described in Q4 defines a function convert(number) that
interprets an integer made only of the digits 0 and 1 (for example, 1100)
as a binary representation and returns its decimal value (for example, 12).
The core idea is to scan the binary digits from right to left, because the
rightmost digit corresponds to the least significant bit with weight 2⁰,
and each step to the left doubles the weight. At the beginning, decimal
is initialized to 0 so it can accumulate the total, place is initialized
to 1 to represent the current bit weight (starting at 2⁰), and n is set
to number so the original input can be processed digit by digit. The loop
runs while n remains positive. In each iteration, remainder ← n mod 10
extracts the rightmost digit of the decimal-form input, which is guaranteed
to be 0 or 1. Then n ← integer part of (n ÷ 10) removes that rightmost
digit, shifting the next binary digit into position for the next iteration.
The conversion step must add the contribution of the current bit to
decimal. Since the current digit is remainder and its weight is
place, the added amount is remainder × place. That means blank A
must be remainder × place, because the update is effectively decimal ←
decimal + (remainder × place). After handling one bit, the next bit to the
left has twice the weight in binary, so place must be updated by doubling
it. Therefore blank B must be 2 × place, corresponding to place ←
place × 2. If you walk through an input like 1100, the first remainder is
0 with place 1, so decimal stays 0 and place becomes 2. The next remainder
is 0 with place 2, so decimal stays 0 and place becomes 4. The next
remainder is 1 with place 4, so decimal becomes 4 and place becomes 8. The
last remainder is 1 with place 8, so decimal becomes 12, and the loop ends
when n becomes 0. With A set to remainder × place and B set
to 2 × place, the completed logic matches choice g, and the function
correctly performs binary-to-decimal conversion.
