#### You may also like 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 ### Repeaters

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.

# Magic Potting Sheds

##### Age 11 to 14 Challenge Level:

We received lots of good solutions to this problem - well done everyone! Many of you spotted that Mr McGregor should put $7$ plants in his potting shed at the beginning, and put $8$ plants in each garden. Well done to Henry from Finton House School, Ruth from Manchester High School for Girls, Liam from Wilbarston School, Mel from Christ Church Grammar School, Rachel from Beecroft Public School in Australia, Yanqing from Devenport High School for Girls and Daniel from Junction City High School for their detailed explanations of how they arrived at the answer.

Henry from Finton House School wrote:

"$1 \times2 \times2 \times2 = 8$.
8 therefore seems likely to be the number in the garden. Let's try it.
The number in the shed at the end must be the number in the garden.
Now what number do we double to get to $8$? It must be $4$.
$4 + 8 = 12$. $12$ divided by $2 = 6$.
$6 + 8 = 14$. $14$ divided by $2 = 7$."

Liam used similar logic:
"Just work backwards from the last garden. Imagine there to be $8$ plants in each garden. (You can't have odd numbers in a garden as the last garden must be double the whole number of plants left after the 2nd garden was planted. I chose $8$ because it's a conveniently sized power of 2.) There must have been $4$ plants left after the 2nd garden was planted so before it was planted there must have been $12$ which is double $6$. $6+8=14$. So Mr McGregor needs to put $14/2$ or $7$ plants in his magic potting shed at the beginning!"

Yanqing and Rachel used algebra. Here is Yanqing's solution:

"First, we make the number of plants put in the shed $n$, and the number planted each night $x$. So by the first morning, the number has doubled to $2n$ in the shed. We plant $x$ of them, leaving $2n-x$ in the shed overnight. By the second morning, we have $2(2n-x)=4n-2x$ in the shed. Planting $x$ of them, we are left with $4n-2x-x=4n-3x$ in the shed. By the third morning, there should be $2(4n-3x)=8n-6x$ plants in the shed. There need to be $x$ plants in the shed, as we need to plant all of them, so $8n-6x=x$ and $8n=7x$.

We can now say that the ratio of $n$ to $x$ is $7:8$, so the smallest values for $n$ and $x$, where they are both positive whole numbers, are obviously $7$ and $8$.

Other numbers which will work are all multiples of $7$."
Rachel also found that $8n= 7x$ and concluded that:
"Now you can see that $8n$ or $7x$ could equal $56$, which makes $n = 7$ and $x = 8$.

This works when you try it out, and if you multiply both numbers by another number, those new numbers work too."
Daniel concluded that:
"If you want to have the same amount of plants in each garden you must start with a multiple of $7$ plants in the shed and each day plant the same multiple of $8$ plants in the garden."

James from C.G.S.B found such a solution:
"Start with $35$ ... then put $40$ in each garden"

And so did Mel from Christ Church Grammar School!
"You start off with $301$ plants in the shed. You put $344$ in each garden."