The article on divisibility tests is very helpful when solving this problem.

- Let the number be 3
1__a__4__b__0__c__92 where a, b, c and d are 5, 6, 7 and 8 in any order.__d__ - 396 is the product 4 x 9 x 11 so if the number is divisible by all of 4, 9 and 11 then it must be divisible by 396.
- Since the last 2 digits are 92, the number must be divisible by 4 (whatever the order of the inserted digits), because 92 is divisible by 4.
- If the digit root is 9, then the number is divisible by 9. The
digit sum is 3 +
*a*+ 1 +*b*+ 4 +*c*+ 0 +*d*+ 9 + 2. Since the order of*a*+*b*+*c*+*d*does not matter then it is always equal to 26. This makes the digit sum 3 + 1 + 4 + 0 + 9 + 2 + 26 which is equal to 45. The digit root is now 4 + 5, which is equal to 9, thus meaning that the number is divisible by 9, no matter what the order of 5, 6, 7, 8. - Now, using the divisibility test for 11:

2 - 9 +*d*- 0 +*c*- 4 +*b*- 1 +*a*- 3 =*a*+*b*+*c*+*d*- 15, and since in any order,*a*+*b*+*c*+*d*= 26 this is equal to 26 - 15 = 11 . This means that the number must be divisible by 11, no matter what the order of the digits 5, 6, 7, 8. - Finally, since the number is divisible by 4, 9 and 11, no matter what the order of the inserted digits, then it must always be divisible by 396! This means that the probability that the answer is a multiple of 396 is equal to 1.

An alternative solution to this was to
try all the numbers one at a time. This was tried successfully
by **Lucy and Sarah**
from Archbishop Sancroft High
School.