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Practice 1

Question 1

Recall the definition of factorial of a number \(n\):

\[n! = n \times (n-1) \times \ldots \times 2 \times 1\]

Use the following pseudocode to calculate the factorial of 10.

Set n to 10
Set f to 1
Repeat until n is 1:
 - Set f to f times n
 - Decrease n by 1
Print f

n = 10
f = 1
while n > 1:
    f = f * n
    n = n - 1 # or n -= 1
print(f)

3628800

Question 2

Find the ‘pseudocode’ box in the notes. Write Python code which implements the program described there.

The value of a loan increases by 1.3% every month, except in December when it is reduced by 100 pounds. How many months will it take for a 1000 pound loan to reach 2000 pounds?

How many months does it take if the monthly interest rate is changed to 0.8%?

# NB the question doesnt specify the starting month, so I have assumed that we start in December. 
# Choosing a different starting month might result in a different answer.

loan = 1000
i = 0
while loan < 2000:
    if i % 12 == 0:
        loan = loan - 100
    else:
        loan = loan * 1.013
    i += 1
print("Number of months:", i)

Number of months: 141

Question 3

Write a program which use the quadratic formula to print the solution to the quadratic equation \(ax^2 + bx + c = 0\), given variables a, b, and c. Your program should print:

The solutions are x = 4 and x = 1

or

The solution is x = 1

or

There are no real solutions

(Hint: first calculate the discriminant \(b^2-4ac\)).

# NB You should test your answer with a variety of values for
# a, b and c to check that it is working correctly.
# At least one test for each of the three type of solution.

a = 1
b = -1
c = -12

discriminant = b**2 - 4 * a * c

if discriminant > 0:
    print("The solutions are x =", (-b + discriminant**0.5)/(2*a), "and x =", (-b - discriminant**0.5)/(2*a))
elif discriminant == 0:
    print("The solution is x =", -b/(2*a))
else:
    print("There are no real solutions")

The solutions are x = 4.0 and x = -3.0

Question 4

Given two integers \(n\) and \(m\) it is possible to perform division-with-remainder by repeatedly subtracting \(m\) from \(n\) until the result is less than \(m\). For example, to calculate 13 divided by 3:

13 - 3 = 10
10 - 3 = 7
7 - 3 = 4
4 - 3 = 1

13 divided by 3 equals 4 remainder 1.

Use the following pseudocode to write a program which performs this calulation.

Set n to 13
Set m to 3
Set i to 0
Repeat while n is greater than m:

  • Subtract m from n

  • Increase i by 1.

Print the result

The result should be displayed in words as above (hint: you’ll need to create an extra variable to store the original value of n).

n = 13
n_original = n
m = 3
i = 0
while n > m:
    n = n - m
    i += 1
print(n_original, "divided by", m, "equals", i, "remainder", n)

13 divided by 3 equals 4 remainder 1