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Adding All Nine

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

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.

A One in Seven Chance

Age 11 to 14
Challenge Level

The remainder when $2^{164}$ is divided by 7 is 4.

Correct solutions came from Alex and Neil of Madras College, St Andrews, who on investigating 2 raised to the power $n$, discovered that the sequence 1,2,4,1,2,4... occurs for increasing values of $n$ which led them to prove their conjectures that:

$$\eqalign{ 2^{3n} \equiv 1 &\text{(mod 7)} \\ 2^{3n+1} \equiv 2 &\text{(mod 7)} \\ 2^{3n+2} \equiv 4 &\text{(mod 7)}}$$

It was Luke, also of Madras College , who went on to investigate 2 raised to the power $n$, (mod $p$) where $p$ is prime. Extending the work of Alex and Neil i.e. 2 raised to the power $n$, (mod 7) which has a period of 3. Luke found that any prime $p$ which can be written in the form $8k+1$ or $8k-1$ has a period less than or equal to $(p-1)/2$, this conclusion holding for $k$ belonging to the set of positive integers and $$2^{p-1} \equiv 1 \text{(mod 7)}$$