Solutions

Solution to Exercise 1.6

pop = 100
i = 0
while pop < 10000:
    pop = pop * 2
    i = i + 1
    print(pop)
    
print("Number of hours:", i)

Solution to Exercise 1.7

pop = 100
i = 0
while pop < 10000:
    pop = pop * 1.02
    i = i + 1
    print(pop)

print("Number of hours:", i)

Solution to Exercise 1.8

pop = 100
i = 0
while pop < 10000:
    pop = pop * 1.02
    i = i + 1
    if i % 24 == 0:
        print(pop)

print("Number of hours:", i)

Solution to Exercise 1.9

half_life = 8.02
r = 0.5 ** (1/half_life)
print('Daily decay rate:', r)

Solution to Exercise 1.10

We would expect that after 8 days the activity would have approximately halved (\(8.35 \times 10^{11}~\mathrm{Bq}\)).

# starting conditions
activity = 1.67 * (10**12)
day = 0

while activity > 10000:
    # activity decreases by the daily decay rate each day
    activity = activity * r
    day = day + 1
    
    # check the activity on day 8
    if (day == 8):
        print('The activity after 8 days is', activity, 'Becquerels')

print('Days taken to reach 10000 Becquerels:', day)

Solution to Exercise 1.11

# starting conditions
activity = 1.67 * (10**12)
day = 0

while activity > 10000:
    # activity decreases by the daily decay rate each day
    activity = activity * r
    day = day + 1

    # print the activity once every 7 days
    if day % 7 == 0:
        print(activity, 'Becquerels')

print('Days taken to reach 10000 Becquerels:', day)

Solution to Exercise 1.12

set n to 5
repeat while n is greater than 1:
    if n is even:
         set n to n/2
    if n is odd:
         set n to 3n + 1
    display n

Solution to Exercise 1.13

n = 5

while (n > 1):
    if (n % 2) == 0:
        # n is even, divide by two
        n = n / 2
    else:
        # n is odd, triple and add one
        n = (n * 3) + 1
    
    print(n)    

Solution to Exercise 1.14

n = 5
collatz_num = 1

# we will change n, so we need to remember what value we started with
start = n

while (n > 1):
    if (n % 2) == 0:
        # n is even, divide by two
        n = n / 2
    else:
        # n is odd, triple and add one
        n = (n * 3) + 1
    
    collatz_num = collatz_num + 1
    
print('The Collatz number of', start, 'is', collatz_num)
        

Solution to Exercise 1.15 (Challenge)

To do this we need to ‘nest’ one while loop (to calculate Collatz numbers) inside another one (to check each if each integer meets the condition).

k = 1
collatz_num = 1

while (collatz_num <= 200): #outer while loop
    #increase k by 1, reset collatz_num
    k = k + 1
    collatz_num = 1

    #remember we don't want to change k itself - or we wouldn't know what numbers we'd already checked
    n = k

    #now find the Collatz number of k
    while (n > 1):
        if (n % 2) == 0:
            # n is even, divide by two
            n = n / 2
        else:
            # n is odd, triple and add one
            n = (n * 3) + 1

        collatz_num = collatz_num + 1

# Once the condition has been met, let's say that we are finished
print('The Collatz number of', k, 'is', collatz_num)
print(k, 'is the smallest number whose Collatz number is greater than 200')