c. CLS Let \( n=321 \) While \( N<>0 \) \( R=n \bmod 2 \) \( S=S+R \) \( N=F I X(n / 10) \) Wend End d. CLS For \( I=1 \) to 5 If \( I \bmod 2<>0 \) then \[ S=S+I \] End if Next I Print S/3 End Questions: i. What happens if we change the function fix with Int? Questions: i. Write the execution result of the above program. ii. Write the numeric functions in the above program. ii. What happens if we change the program line? \( I \bmod 2<>0 \) to \( I \bmod 2=0 \) ?

If we replace the `FIX` function with `INT`, rounding down would still lead to the same integer result for positive numbers, like in your code with \( n=321 \). However, if \( n \) were negative, `INT` truncates towards negative infinity, giving different outcomes than `FIX`, which simply discards the decimal part. The execution result of the program would reveal that the value \( S \) accumulates the remainders of \( n \) divided by 2 until \( N \) reaches 0, essentially counting the number of 1s in the binary representation of \( n \). For your input \( n=321 \), \( S \) would end up at 5, reflecting the binary form of \( 321 \) being \( 101000001 \) which contains five 1s. Changing the condition from \( I \bmod 2<>0 \) to \( I \bmod 2=0 \) will result in \( S \) summing only even values of \( I \) rather than odd ones. This means \( S \) would only include 2 and 4, leading to a significantly smaller total output when printed as \( S/3 \). You’d see a different numeric result, reflecting the alternating sequence of odd and even integers!