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Adding All Nine

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

Double Digit

Choose two digits and arrange them to make two double-digit numbers. Now add your double-digit numbers. Now add your single digit numbers. Divide your double-digit answer by your single-digit answer. Try lots of examples. What happens? Can you explain it?


Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

More Magic Potting Sheds

Age 11 to 14
Challenge Level

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.