Prove that there exists no natural number such that shifting its first digit to the end, multiplies it by 5, 6, or 8.

For each case, if the first number is shifted to the last position, the number of digits remains the same.

The first number must be a 1 for each case of 5, 6, or 8. Otherwise when multiplied by 5, 6, or 8, the new number would be one digit longer than the original.

Using 1 as the first digit, if the 1 is moved to the last position then it would automatically not be divisible by 5, 6 or 8.  (At least in base 10).

Now the fun of searching other bases . . .

Posted by Leming on 2007-03-26 13:34:01