# Divisible digits

Can you find the missing digits, given that the number is divisible by 3, 4, 5 and 6?

I wrote down a 4-digit number that was divisible by 3, 4, 5 and 6, but I spilt a cup of tea on it and can only see the first two digits.

The first two digits are 95 (in that order). What were the last two?

*This problem is adapted from the World Mathematics Championships*

**Answer**: 9540

**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.