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 (rightmost) 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?
If we denote the other digits of the number (than the initial or trailing 6) as "string", we seek a number 6string = 4 x string6. That is string6 = 6string / 4.
Similar to one of the the solutions of Move the 2, Double the Number, we start the long division:
1
4)6
and note
that we can bring down the digit 1 in the quotient to be the digit after the 6 in the dividend, and continue the long division:
15
4)61
4
21
Then we can bring down the 5 the same way
and continue:
153846
4)615384
4
21
20
15
12
33
32
18
16
24
24
0
That's the first point where we can stop, as the remainder is zero.
But we also could choose to continue, and if we do we must bring down the 6 to be the next digit after the 4, and thus it starts again as from the beginning. So any such number must be the string 153846, either once or repeated any number of times.
So the first (smallest) such number is 153,846 and the second is 153,846,153,846 and the tenth is 153,846,153,846,153,846,153,846,153,846,
153,846,153,846,153,846,153,846,153,846.

Posted by Charlie
on 20030904 14:47:24 