Let the user enter a limit value .Create a loop that will
calculate powers of 2 (i.e., 1, 2, 4, 8, 16 …) and place them into
the listbox. You should stop the loop as soon as the calculated
power of 2 exceeds the user’s limit value. If the user entered 10,
you would end up with 1, 2, 4, 8,
and then stop. If the user entered 16, you would end up with 1, 2,
4, 8, 16 in the listbox
and stop.
As understood, there are three questions. Let's tackle them one by one. Since no language is specified, general algorithm based style will be used to explain.
ANSWER 1
NOTE : The list is needed to be dynamic because the size of the list is not known. For this purpose, linked ist in C can be used, JAVA has ArrayList and python has built-in list data structure. Nevertheless, it is assumed that we have a readymade data sturucture like that in JAVA or python.
Algorithm:
1. Initialise the empty list L, and a variable to store sum, Sum=0.
2. Generate a random number and take it's modulus 11, say the result is R and add it to the Sum and append it to the list L.
3. If Sum is less than 100, repeat step 2, else continue to step 4.
4. Print all the values one by one from the starting of the list.
Python Code :

ANSWER 2:
This question also needs dynamic lists.
Algorithm:
1. Initialise the empty list L, and a variable to store sum, Sum=0.
2. Generate a random number and take it's modulus 11, say the result is R.
3. If Sum+R is less than 100, add R to the Sum and append R to the list then repeat step 2, else continue to step 4.
4. Print Sum and R value and stop.
Python Code;

ANSWER 3:
This question can be done without dynamic list, by finding out the number of elements required beforehand. But here solution is given considering dynamic list.
Algorithm:
1. Take user input in a variable Inp.
2. Initialize variable Cnt from 1.
3. If Inp is less than Cnt, stop here else continue to step 4.
4. Add Cnt to list L and multiply Cnt by 2.
5. Check if Cnt is less than or equal to Inp. If yes then repeat step 4 else go to Step 6.
6. Print all the values of L starting from the first one.
Python Code:

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