Briefly

The multi-divisible number

The multi-divisible number

See if you are able to find a nine-digit number that meets the following conditions:

  • All digits 1 through 9 must appear only once
  • The number must be divisible by 9
  • If we delete the last digit on the right, the number must be divisible by 8
  • If we delete the last two digits on the right, the number must be divisible by 7
  • If we delete the last three digits on the right, the number must be divisible by 6
  • If we delete the last four digits on the right, the number must be divisible by 5
  • If we delete the last five digits on the right, the number must be divisible by 4
  • If we remove the last six digits from the right, the number must be divisible by 3
  • If we delete the last seven digits on the right, the number must be divisible by 2
  • If we delete the last eight digits on the right, the number must be divisible by 1

Solution

Let's call the ABCDEFGHI number where each letter will represent a different digit. It is clear that the digits B, D, F and H must be even since they correspond to the last digit of the numbers that must be divisible by even numbers (2, 4, 6 and 8). The rest will therefore be odd digits since we know that you must include all the numbers from 1 to 9.

Since ABCDE is divisible by 5, we know that E has to be equal to 5.

Since ABCD is divisible by 4, it will also be fulfilled that CD will be divisible by 4 and GH will be divisible by 8 (since FGH will be divisible by 8 and F is even).

Because C and G are odd, D and H must be 2 and 6 but not necessarily, in this order.

We know that ABC is divisible by 3, that ABCDEF is divisible by 6 and therefore also by 3 and that ABCDEFGHI is divisible by 9 and therefore also by 3 so that A + B + C, D + E + will be fulfilled F and G + H + I are divisible by three.

If we assume for example D = 2, then it would be true that F = 8, H = 6 and B = 4. A + 4 + C is divisible by 3, therefore, A and C must be 1 and 7 or vice versa and G and I must be 3 and 9 or vice versa. GH is divisible by 8, therefore it must be agreed that G = 9 and from the previous conclusion we obtain that I = 3. In this case the possible numbers 1472589 and 7412589 are not divisible by 7. Therefore it must be met what D = 6 Where do we deduce that F = 4, H = 2, B = 8.

G + 2 is divisible by 8, therefore, G can only be 7 or 3.

A + 8 + C is divisible by 3 and therefore the values ​​of A and C must be one 1 or 7 and the other 3 or 9.

If we assume for example G = 3, then A or C must be 9 and the other must be 1 or 7. But none of the numbers 1896543, 7896543, 9816543 and 9876543 are divisible by 7. Therefore G = 7 and then A or C must be equal to 1 and the other 3 or 9. Of the possible numbers 1836547, 1896547, 3816547 and 9816547, 3816547 only the latter is divisible by 7 (the quotient is 545221). Thus, The number we are looking for is 381654729.

You can find more information about polydivisible numbers on Wikipedia.