In Move the 2, Double the Number, we found a number that ended in two, for which moving the two to the beginning of the number doubled its value.

For this problem, we have a number with 6 as the last (right-most) digit.
If we erase the 6 and put it on the left end of the number (for example, 936 would become 693), then we have a number four times our original number (we see that 936 doesn't work, of course).

What is the smallest number that fits this condition?

What is the second smallest number that works?

What is the tenth smallest number that works?

The nth smallest positive integer that fits the condition is
= (20/13*(10^(6n-1) - 4)) + 6

1st number = (20/13*(10⁵ - 4)) + 6 =
153846

2nd number = (20/13*(10¹¹ - 4)) + 6 =
153846153846

10th number = (20/13*(10⁵⁹ - 4)) + 6 =
153846153846153846153846153846153846153846153846153846153846

Edited on November 28, 2007, 9:32 am

Posted by Dej Mar on 2007-11-28 09:29:14