Strange number (2)

Let x be a number satisfying the condition of the problem.

Properties of the first digit:

Since 6x must be, along with x, a six-digit integer, the first digit of x must be 1 and the following digit cannot exceed 6. Otherwise, 6x will be a 7 digit number. So we can conclude that:

(1) The leading digits of the numbers x, 2x, 3x, 4x, 5x, and 6x are all different and, as a result, must comprise all the digits contained in the integer x (each of these digits must appear in the original number x).

(2) All the digits of x are different.

(3) Since the first digit of x cannot be 0, otherwise it will be a 5 digit number, therefore none of the digits is 0.

Properties of the last digit:

(1) Since there is no 0 in the number, the last digit of 5x can only be 5. That means the last digit of x must be odd. The only possibilities for the last digit are (1, 3, 5, 7 or 9).

(2) Since 1 is already taken in the first position, the remaining possibilities for last digit are (3, 5, 7 or 9).

(3) Since the last digit of x cannot be 5, otherwise 2x will end with 0, the remaining possibilities for last digit are (3, 7 or 9).

(4) If the last digit of x was 3, the last digit of x, 2x, 3x, 4x, 5x and 6x should be 3, 6, 9, 2, 5 and 8 respectively. This is impossible because these are already 6 possibilities without the 1 which will make x a 7 digit number. This should be excluded. Therefore the remaining possibilities for last digit are (7 or 9).

(5) Similarly, if the last digit of x was 9, the last digit of x, 2x, 3x, 4x, 5x and 6x should be 9, 8, 7, 6, 5 and 4 respectively. This is impossible because these are already 6 possibilities without the 1 which will make x a 7 digit number. This should be excluded.

(6) The only possibility for the last digit is 7. In this case the last digit of x, 2x, 3x, 4x, 5x and 6x should be 7, 4, 1, 8, 5 and 2 respectively.

We can now conclude that the 6 digits of x are 1, 2, 4, 5, 7 and 8 in a certain order. We can also deduce that the numbers x, 2x, 3x, 4x, 5x, and 6x are in the following form (asterisks to represent unknown digits):

x.1 = 1 * * * * 7

x·2 = 2 * * * * 4

x·3 = 4 * * * * 1

x·4 = 5 * * * * 8

x·5 = 7 * * * * 5

x·6 = 8 * * * * 2

Also we should note that, not only must each row contain (on the right) all six distinct digits 1, 2, 4, 5, 7, and 8, but each column must contain these six distinct digits, in some order. Because if we suppose that x.2 and x·5 have the same digit a in a certain position, say in the third position we should have the difference x·5 - x·2 = x·3 will be a six-digit number, either the digit 0 or the digit 9 will stand in its third position (since we can take, at most, a unit carry-over into this place for the subtraction). But this is an impossibility, since we already know that the number x· 3 will contain neither a 0 nor a 9 as one of its digits. Therefore, in the above display of x.1; x·2; x.3; x.4; x.5 and x.6, the sum of the digits in any column is 1 + 2 + 4 + 5 + 7 + 8 = 27. We can therefore add the right member of this display and obtain x·21 = 2,999,997

Solving this equation gives x = 142,857, which is the integer sought. As a check we have:

x.1 = 142,857

x.2 = 285,714

x.3 = 428,571

x.4 = 571,428

x.5 = 714,285

x.6 = 857,142

$Answer = \boxed{142857}$