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

• Let the number be 3 a 1 b 4 c 0 d 92 where a, b, c and d are 5, 6, 7 and 8 in any order.
• 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.