More magic potting sheds
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
Age
11 to 16
Challenge level
Problem
This problem follows on from Magic Potting Sheds
After a year of successful gardening using his magic doubling shed (introduced in Magic Potting Sheds), Mr McGregor buys a new shed that trebles the number of plants in it each night. Use the interactivity to investigate how many plants he needs this time to get the same number in each garden. What is the smallest number of plants he could use?
Move the slider at the bottom to choose the Magic Multiplier for the shed. Type in the input box or use the slider below it to choose how many plants to buy.
Can you predict how many plants he would need on the first day and how many he should plant each day if he bought a new shed that quadruples the number of plants in it each night?
Use the interactivity to test your prediction.
Mr McGregor is so successful that he decides to plant more gardens. He can still only plant one garden each day.
Use the interactivity to change the number of gardens and investigate how many plants he should use for each of the different potting sheds.
What do you find?
Can you find a general rule?
Can you explain why your rule works?
Unfortunately, Mr McGregor suffers an attack from evil magic slugs that eat half of the plants in his (non-magic) potting shed each night. He still wishes to plant the same number of plants in each garden.
How many plants does he need on the first day this time, and how many should he plant each day? (Remember that he can only plant whole numbers of plants!)
Getting Started
You may like to start by having a go at the Magic Potting Sheds problem.
It may be useful to keep a record of your trials.
Try working backwards from the number of plants in the last garden (remembering that you must always have whole numbers of plants in the gardens and shed).
Describing the different stages algebraically may be helpful.
Student Solutions
Well done to Liam from Wilbarston School and Ruth from Manchester High School for Girls for sending us their work on this problem.
Here is Liam's work on the problem for a Magic Growth Factor of 3. He used some ideas from his solution to Magic Potting Sheds to get started.
Let's start by saying there are going to be $27$ plants in each field. In Magic Potting Sheds I used $8$ which is $2^3,$ so for magical growth factor $3$ I'll try $3^3.$ ($9$ doesn't work, because the morning of the day Mr McGregor planted the 1st garden he should have had 13 plants - a number which is not divisible by $3$.)
Now, working back from the last garden, there must have been $9$ after Mr McGregor planted his 2nd garden. Now as there has to be the same number of plants in each garden, before he planted his 2nd garden he had to have had $36$ ($9+27$) plants... which means that he had $12$ plants after planting $27$ in his 1st garden making $39$ altogether... which is $3$ times $13$.
You must start with $13$ in order to get $27$ in each garden.
Liam also used the same method to see that
for a magic growth factor of $4$, Mr McGregor would need to plant
$64$ plants in each garden, and would need $21$ plants at the
beginning.
Ruth, from Manchester High School for Girls, used algebra to find what Mr McGregor needs to do in each situation. Well done!
If the shed multiplies the number of plants by $x$ every night, you start with $x^2 + x + 1$ plants and plant $x^3$ every day. After the first night you have $x^3+x^2+x$ and plant $x^3$, leaving $x^2 + x$. After the second night you have $x^3+x^2$ then plant $x^3$ so you have $x^2$ which becomes $x^3$ to plant on the third day.
If you have $n$ nights instead of 3, you start with $x^{n-1} + x^{n-2} + \ldots + x^2 + x + 1$ plants and plant $x^n$ every day.
After the first day you have $x^n + x^{n-1} + \ldots + x^3 + x^2 + x$ and plant $x^n$.
Every night each term's exponent is increased by $1$ and when you plant $x^n$ plants you remove $1$ term until on the $n^{th}$ day the term that started as $1$ is $x^n$ and the last lot of plants left.
When the numbers of plants halves each night, the smallest solution is to plant 1 plant each day. You need $2^{1}+2^{2}+2^{3}=14$ plants. Each night the exponent decreases and when you plant you get rid of a term. If you have $n$ nights you start with $2^{1}+2^{2}+2^{3}+ \ldots +2^{n-2}+2^{n-1} +2^{n}=2^{n+1}-2$.
When the number of plants is divided by $y$ you start with $y^1+y^{2}+y^{3}+ \ldots + y^{n-2}+y^{n-1}+y^{n}=\frac{y^{n+1}-y}{y-1}$ and plant $1$ each day.
Ruth also extended her solution one step beyond what we had asked.
When the number of plants is multiplied by $\frac{x}{y}$, you need $x^{n-1} y^{1} + x^{n-2} y^{2}+\ldots + x^{2} y^{n-2} + x^{1} y^{n-1}+ x^{0}y^{n}$ plants for $n$ nights and plant $x^{n}$ each night.
Teachers' Resources
This problem requires students to work systematically and
challenges them to arrive at further generalisations. It is an
excellent exemplar of a context where a full understanding of one
problem is really useful when applied to the next problem. It is
useful as an object lesson in mathematical process skills.
The problem can be tackled on paper but access to the interactivity will speed up the process of exploring different possibilities. It is important to have paper available for students to record the results of the different trials.
Teacher's Notes
Start with Magic Potting Sheds
Once students have found several answers to this problem, bring the class together.Suggest that if they can understand why the solutions involve multiples of $7$ and $8$ they will find it a lot easier to solve similar problems.
Ask students if they can explain why the solutions are multiples of $7$ and $8$.When I have used this with students, I've found that a particularly effective intervention at this point is to offer the following way of picturing the situation:
Imagine Mr McGregor places his $7$ plants on three shelves:
$4$ on the bottom shelf, ready for planting the next day
$2$ on the middle shelf, ready for planting the day after
$1$ on the top shelf, ready for planting on the last day.
On the first night they become $8, 4$ and $2$ and the $8$ are planted in the first garden.
That leaves $4$ on the middle shelf and $2$ on the top shelf.
The following night they become $8$ and $4$, and the $8$ are planted in the second garden.
That leaves $4$ on the top shelf.
On the last night the $4$ become $8$, and they are planted in the third garden.
So $7$ is significant because it is the sum of $1, 2$ and $4$ (the smallest triple of numbers that have the required doubling relationships).
Equally, Mr McGregor could have placed $5, 10$ and $20$ plants on his three shelves (multiples of $\{ 1, 2, 4 \}$):
The $20$ will become $40$ after one night.
The $10$ will become $40$ after two nights (double and double again).
The $5$ will become $40$ after three nights (double and double and double again).
Students can now be set to work on More Magic Potting Sheds, using this or other insights to help them along the way. Encouraging students to reflect on the solutions to the original problem may help them tackle the follow-up questions in a more informed way.
The intention is to avoid the situation in which students just sit at the computers typing any value in until they eventually hit upon the answer. It may be a good idea to set students working on the follow-up questions with paper and pencil, and only allow them to move to the computers once they have clear ideas about what they should try.