Question

the block of code below is a small portion of a program. which of the following starting values of n would make this portion of a program repeat the most amount of times? repeat until n > 10 if n * 2 < 15 n ← n + 3 else n ← n + 1

Explanation:

Step1: Analyze the loop condition

The loop repeats until $n>10$. Inside the loop, if $n * 2<15$ (i.e., $n < 7.5$), $n$ is incremented by 3, otherwise by 1.

Step2: Test each option

• If $n = 1$:
• First iteration: Since $1*2 = 2<15$, $n$ becomes $1 + 3=4$.
• Second iteration: Since $4*2 = 8<15$, $n$ becomes $4 + 3 = 7$.
• Third iteration: Since $7*2=14<15$, $n$ becomes $7 + 3 = 10$.
• Fourth iteration: Since $10*2 = 20>15$, $n$ becomes $10+1 = 11$, and the loop stops.
• If $n = 2$:
• First iteration: Since $2*2 = 4<15$, $n$ becomes $2 + 3=5$.
• Second iteration: Since $5*2 = 10<15$, $n$ becomes $5 + 3 = 8$.
• Third iteration: Since $8*2=16>15$, $n$ becomes $8 + 1=9$.
• Fourth iteration: Since $9*2 = 18>15$, $n$ becomes $9+1 = 10$.
• Fifth iteration: Since $10*2 = 20>15$, $n$ becomes $10 + 1=11$, and the loop stops.
• If $n = 3$:
• First iteration: Since $3*2 = 6<15$, $n$ becomes $3+3 = 6$.
• Second iteration: Since $6*2 = 12<15$, $n$ becomes $6 + 3 = 9$.
• Third iteration: Since $9*2=18>15$, $n$ becomes $9 + 1=10$.
• Fourth iteration: Since $10*2 = 20>15$, $n$ becomes $10+1 = 11$, and the loop stops.
• If $n = 4$:
• First iteration: Since $4*2 = 8<15$, $n$ becomes $4 + 3=7$.
• Second iteration: Since $7*2 = 14<15$, $n$ becomes $7 + 3 = 10$.
• Third iteration: Since $10*2=20>15$, $n$ becomes $10 + 1=11$, and the loop stops.

We can see that starting with $n = 1$ makes the loop repeat the most number of times among the given options.

Answer:

1