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?
Take any four digit number. Move the first digit to the 'back of
the queue' and move the rest along. Now add your two numbers. What
properties do your answers always have?
Choose any four consecutive even numbers. Multiply the two middle
numbers together. Multiply the first and last numbers. Now subtract
your second answer from the first. Try it with your own numbers.
Why is the answer always 8?
Make a set of numbers that use all the digits from $1$ to $9$,
once and once only.
For instance, we could choose:
$638, 92, 571$ and $4$
Add them up:
$638 + 92 + 571 + 4 = 1305$
$1305$ is divisible by $9$ (it is $145\times 9$)
(use a calculator to check this if you do not know yet how to
divide by $9$)
Add each of the digits in the number $1305$ .
What is their sum?
Or, perhaps we could choose:
$921, 4357$ and $68$
$921 + 4357 + 68 = 5346$
$5346$ is divisible by $9$ (it is $594 \times 9$)
Now try some other possibilities for yourself!
I wonder what happens if we use all $10$ digits from $0$ to $9$,
once and once only?
Try some for yourself!
What do you think would happen if we used the eight digits from
$1$ to $8$?
Test your hypothesis by trying some possibilities for
Were you correct?
Is there a pattern beginning to emerge? Do you have theory that
might explain what is happening?
Try some different sets of digits for yourself!