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Counting Factors

Is there an efficient way to work out how many factors a large number has?

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.


Can you select the missing digit(s) to find the largest multiple?

Digital Division

Age 11 to 14 Short
Challenge Level

A number is a multiple of 6 precisely when it is both a multiple of 2 and of 3. To be a multiple of 2, it will need to end with an even digit; i.e. 0 or 2. To be a multiple of 3, the sum of the digits has to be a multiple of 3.

If it ends with 0, the sum of the other two digits must be a multiple of 3; and only $3 = 1 + 2$ or $6 = 1 + 5$ are possible. That gives the numbers 120, 210, 150, 510.
If it ends with 2, the sum of the others must be $1 = 0 + 1$ or $4 = 1 + 3$. That gives 102, 132 and 312.

Hence 7 of these numbers are divisible by 6.
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.