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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.

Big Powers

Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.

Add to 200

Age 11 to 14
Challenge Level

Lucy and Rebecca from Preston Academy investigated when the total is even or odd:

In order to know if the total was going to be odd or even, we drew a grid (like the one shown below) and began writing out all the different possibilities. In this grid, $O$ stands for odd and $E$ stands for even:

$X$ $E$
$E$ $E$

We put an $X$ in the top, left-hand box as this is the ‘tens’ unit whenever it appears, so it does not affect whether the total is odd or even. We then filled in all the other boxes with an $E$. From looking at the rules for adding odd and even numbers, we found out that when all the boxes (except the top, left) have an even digit in them, the total will always be even.

They tried different combinations of odd and even numbers in their grids, and found:

if you have an odd number in the bottom, right hand box and then use one odd number and one even number for the other two boxes (top right and bottom left), it does not matter where they go as you will still get the same result.

Julian from the British School Manila in the Philippines used a more algebraic approach:

Let the four digits be $a$, $b$, $c$, $d$, arranged in the grid as follows:

$a$ $b$
$c$ $d$

Therefore the first horizontal number is $10a+b$, and the second horizontal
number would be $10c+d$.

The two vertical numbers are $10a+c$ and $10b+d$.

The final sum of these four numbers would be $10a+b+10c+d+10a+c+10b+d$

Note that $10a+d$ is the two-digit number with first digit $a$ and second digit $d$.

$2(10a+d)$ will always be even. Even+Even=Even and Even+Odd=Odd, so if $11(b+c)$ is
even the total will be even, and if $11(b+c)$ is odd the total will be odd.
Now note $11$ times an odd number is an odd number, and $11$ times an even number is an even number, so if $b+c$ is even the total will be even, and if $b+c$ is odd the total will be odd.


The quick way to tell if it's even or odd is by adding the top right and bottom left number.

Lots of people, including George and Alex from Malet Lambert School, Ariz from Warren Primary School and Hazel Smith from Kings Norton Girls' School found some of the solutions making $200$:


$8$ $1$
$1$ $9$
$4$ $5$
$5$ $5$
$3$ $6$
$6$ $4$

Mitchell from St Mary's Greensborough found another solution:

$4$ $3$
$7$ $5$

Well done everyone!

Julian continued his algebraic approach to try and calculate the number of ways to make $200$:

Remember we have calculated that for numbers $a$, $b$, $c$, $d$, our total will be $2(10a+d)+11(b+c)$.

Suppose we are looking for a target of $200$: $2(10a+d)+11(b+c) = 200$.

Because $200$ is even, $b+c$ must be even (by the criterion from the first part of the question)

We know that $10a+d$ is a number between $0$ and $99$, so $2(10a+d)$ is between $0$ and $198$.

If $2(10a+d)+11(b+c) = 200$, by rearranging we can see $2(10a+d) = 200-11(b+c)$.  So we know that $200-11(b+c)$ is between $0$ and $198$.

By taking multiples of $11$ away from $200$, we can see that this restricts possible values for $b+c$ are $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $11$, $12$, $13$, $14$, $15$, $16$, $17$, and $18$. 

But we also know that $b+c$ must be even, so we can only have $b+c$ being $2$, $4$, $6$, $8$, $10$, $12$, $14$, $16$, and $18$.

Because $b+c$ determines the value of $10a+d$ and therefore $a$ and $d$, the number of different solutions where, e.g. $b+c =2$ are the number of different pairs of numbers we can find which add to $2$.

This means that there are a lot of solutions altogether...!


Julian also thought about the different totals between $0$ and $396$ he could make:

Because our total is $2(10a+d)+11(b+c)$, all numbers in the form $2x+11y$ can be made ($0\leq x \leq 99$, $0 \leq y \leq 18$).

The only sums that cannot be made are $1$, $3$, $5$, $7$,
and $9$.

Method: If the number is even, find any multiple of $11$ that is also
even (or if the number is odd, find a multiple of $11$ that is also odd), and is less than or equal to $198$ ($18 \times 11$) but bigger than your number. 
Take the difference between your number and this multiple of $11$, halve it and label the first digit of this number $a$, and the second $d$.

Divide your multiple of $1$1 by $11$ and label this $b+c$. Find any pair of numbers $b$ and $c$
that adds up to your value for $b+c$.
Finally put the numbers in the table like follows:

$a$ $b$
$c$ $d$

and you have a solution.

Well done!

Thank you to everyone who submitted a solution to this problem.