### 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.

### Dozens

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

# Digital Division

##### Age 11 to 14 ShortChallenge 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.