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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.