Layla's calculator

n = A = 11.

While i < 11 will make the loop runs 10 times (from 0 to 10).

(10*3) + 3 = 33;

answer = 0+3; i turns 1;

answer = 3+3; i turns 2;

answer = 6+3; i turns 3;

...

answer = 27+3 and i turns 9;

answer = 30+3 and i turns 10;

Here we have (i < n), or (10 < 11) == (true). Loop breaks.

So, answer returns 33 when n = 11.!

Hope it helps!

Here,the loop runs n times and the value of n is passed into the function .Every time the loop runs, the variable answer is incremented 3 times which shows that this loop basically gives the table of 3 Hence,if the function is returning 33 it means that the loop must run 11 times as 11*3=33 So, the value of n should be 11 so,answer is 11.

when n=11, while loop will be iterated 11 times (as i<n) to return 33. If condition is defined as(i<=n) then, for result 33 n should be 10.

It's looping problem:

if $n=11$

So, answer=answer+3 looping from n=0 until n=10.

Since answer (0)=0, this program is recursive until n=10.

So, 3+3+...+3=33, where A=11

This is called as iteration in programming. Here it will be iterated for 11 times starting from 0 to 10. for each iteration answer is added with 3 so 11 times which can be simplified as (11*3 = 33)

The counter adds $1$ and the function adds $3$ . This is done from $i = 0$ to $i = n - 1$ , that is: a total of $n$ times. Therefore, the program adds 3 a total of n times, i.e. finds $3 \cdot n$ . Since we have $3n = 33$ , the original argument must have been $\boxed {11}$ .

Nice problems, you can solve with some modification of the code like this:

# Layla's calculator - in Python

def calculate n times_3(n):

i = 0

answer = 0

while (i < n):

    answer = answer + 3

    i = i + 1

return answer

now you can call to get A :

for j in range(0,20):

if calculate_n_times_3(j) == 33:

    print(j)

Using loop, from 0 to 20 (you can edit that, but, I can get answer just stopped at 11). So, the answer about number A is 11 .