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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.

Forwards Add Backwards

Age 11 to 14
Challenge Level

Lots of people submitted solutions to this problem - well done, everyone!

Matthew from West Island School in Hong Kong found the other two sets of numbers adding to $726$:

726 can be made by adding 165, 264 and 363 with their reversals.

Hank and Vacha from Elm Park School in New Zealand gave these answers too, along with a great explanation for how they worked them out:

We knew that the units had to add up to $6$ because we couldn't use decimals. The only ways to make $6$ were $3+3$,  $5+1$ and $2+4$ ($2+4$ leads to the sum given in the question).

So the first answer has to be something like this: $3?3 + 3?3 = 726$.
The only ways to make $2$ in the tens column are $1 +1$ and  $6+6$. It can't be $1+1$ because then the total will be $626$, so the number in the tens place must be $6$. Then you
repeat this step with $1$ and $5$ in the units column to find $561+165=726$.

The same method also works for finding the three ways to do this for $707$ and $766$. Ashlyn from St. Stephen's School Carramar gave this explanation for $707$:

$3$ ways of forming $707$ by adding a number to its reversal:


The $1^{st}$ digit and $3^{rd}$ digit of the number must total $7$ (the $3^{rd}$ digit of

$0$ cannot be used as a $1^{st}$ or $3^{rd}$ digit as you need both the number and its reversal to have three digits.

The $2^{nd}$ digit in any three digit number and its reversal must be the same.

We can find which numbers add to give $7$:


We know we can't use $0+7$, so the other three combinations must be the $1^{st}$
and $3^{rd}$ digits.

Now we use a system of trial and error to find the $2^{nd}$ digit. We find that $0$ is the only one that fits.

Therefore the $3$ combinations must be:


Catherine and Charlotte from Culford School found the three numbers which made $766$ when added to their reversal:


Beth and Emily from Culford school successfully found the $10$ numbers between $700$ and $800$ which can be written as the sum of a number and its reversal, and spotted some patterns:

$706$, $707$, $726$, $727$, $747$, $746$, $767$, $766$, $787$, $786$.
They all have even numbers in the middle. They all end in either $6$ or $7$.

Michael Presland from Sevenoaks School added this explanation for the pattern:

For the numbers ending in $7$ the numbers on the outside (of the numbers
you are adding together) must be $3$ and $4$, $6$ and $1$, or $5$ and $2$.
The number in the middle must be less than $5$ so there is no carry over.

For the numbers ending in $6$, we found that the number in the middle of
the two numbers you are adding together must be $5$ or more, so as to make the digit in the hundreds column be $7$ (by carrying one over) while the digit in the units column is six.

Some people tried using an algebraic method to explain the pattern.

Let's say that our three-digit number is written $ABC$. We can express this algebraically as $100A+10B+C$.
Then the reversal of our number is $CBA$, which we can express as $100C+10B+A$. Adding these together gives $101(A+C) + 20B$.

If $A+C$ were greater than $9$, our total would be bigger than $1000$: as we are looking for numbers between $700$ and $800$, $A+C$ must be a single digit.

Let's think about the sum:

   $A$ $B$ $C$
+ $C$ $B$ $A$

The units value in this total is $A+C$, with no carry over.

In the tens column, we calculate $B+B$. If $B$ is less than $5$, the tens value is $2B$ with no carry over. If $B$ is between $5$ and $9$, the tens value is the last digit of $2B$ with a carry over of $1$.

The hundreds value in this total is either $A+C$ (if there is no carry over from the tens column), or $A+C+1$ (if there is a carry over from the tens column).

We want the hundreds value in this total to always be $7$, as we are looking for numbers between $700$ and $800$. So either $A+C=6$ or $A+C=7$. This means the units value in the total is always $6$ or $7$. 

Whatever $B$ is, the last digit of $2B$ will always be even. So the second digit in our total will always be $0$, $2$, $4$, $6$, or $8$.

Lots of people correctly spotted that there are $10$ numbers between $300$ and $400$ which can be written as the sum of a three-digit number and its reversal, and $10$ numbers between $800$ and $900$ with this property. Will, Lew, and Nick from All Saints Junior School found the numbers:

Between $300$ and $400$ :  $363$,$383$,$323$,$343$,$303$,$322$,$302$,$342$,$362$,$382$
Between $800$ and $900$ :  $808$,$828$,$848$,$868$,$888$,$807$,$827$,$847$,$867$,$887$.

Well done to everybody who submitted a solution to this problem!