### Counting Factors

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

### Repeaters

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.

### Helen's Conjecture

Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?

# Divisible Digits

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

Using divisibility tests
9  5   ___   ___
9  5   ___  (0 or 5) (since divisible by 5)
9  5   ___   0 (divisible by 6, so even)

Divisible by 3 $\Rightarrow$ sum of digits a multiple of 3
9 + 5 = 14, so this digit can be 1, 4 or 7
9510, 9540 or 9570

Divisible by 4 $\Rightarrow$ last 2 digits divisible by 4
9540

Note: You can apply the divisibility tests in a different order, but some orders will take longer than others!

Finding a larger number that the number must be divisible by
The number is divisible by 3, 4, 5 and 6, so it is divisible by 60 (lowest common multiple).

The number is less than 9600. 96 = 60 + 36, so 96 is a multiple of 6, so 960 is a multiple of 60 and therefore 9600 is a multiple of 60.

Multiples of 60: 9600, 9540, 9480

Only 9540 begins 95.

You can find more short problems, arranged by curriculum topic, in our short problems collection.