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.