You may also like

problem icon

N000ughty Thoughts

How many noughts are at the end of these giant numbers?

problem icon

Mod 3

Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.

problem icon

Novemberish

a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.

396

Stage: 4 Challenge Level: Challenge Level:1

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.