We are asked to find the output of two pseudo code snippets. The first one is a simple loop that prints the value of $n$ and decrements it. The second one involves updating variables $m$ and $n$, comparing them and printing one of them.

1. Problem Description

We are asked to find the output of two pseudo code snippets. The first one is a simple loop that prints the value of $n$ and decrements it. The second one involves updating variables $m$ and $n$, comparing them and printing one of them.

2. Solution Steps

Problem 33:

The loop `forn = 3 tol do` means the loop iterates from 3 down to

1. Inside the loop, we print the value of $n$, then decrement it: $n = n - 1$.

Iteration 1: $n = 3$. Print $n$ (3). $n$ becomes $3 - 1 = 2$.

Iteration 2: $n = 2$. Print $n$ (2). $n$ becomes $2 - 1 = 1$.

Iteration 3: $n = 1$. Print $n$ (1). $n$ becomes $1 - 1 = 0$.

The loop stops because $n$ is now less than

1. The output is 3 2

1.

Problem 34:

We have $m = 9$ and $n = 6$.

$m = m + 1$, so $m = 9 + 1 = 10$.

$n = n - 1$, so $n = 6 - 1 = 5$.

$m = m + n$, so $m = 10 + 5 = 15$.

The condition is `if (m>n)`. Since $15 > 5$, the condition is true.

We print $m$, which is

1

5.

3. Final Answer

Problem 33: (2) 3 2 1

Problem 34: (4) 15