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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.

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

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 $4$

$638 + 92 + 571 + 4 = 1305$

$1305$ 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 yourself!

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!