Eugene Rabii wrote:I have to say, this is impressive!

I have reached the same code:

x = ( 2*(10^n) - 4 ) / 19

And it was a pleasure getting here!

Cheers, Eugene

-4? -2 surely?

My thinking runs this way:

2 * (10^n) + (x-2)/10 = 2x

2 * (10^(n+1)) + x - 2 = 20x

2 * (10^(n+1)) - 2 = 19x

x = (2 * (10^(n+1)) - 2 ) / 19

So, we just look for a value of n that makes x an integer ending with 2. The n vs (n+1) doesn't matter, but the -4 vs -2 will produce different answers.

OK, code time:

Results:

2, 10, 8

3, 105, 3

4, 1052, 10

5, 10526, 4

6, 105263, 1

7, 1052631, 9

8, 10526315, 13

9, 105263157, 15

10, 1052631578, 16

11, 10526315789, 7

12, 105263157894, 12

13, 1052631578947, 5

14, 10526315789473, 11

15, 105263157894736, 14

16, 1052631578947368, 6

17, 10526315789473684, 2

18, 105263157894736842, 0

So the answer is 105263157894736842. The check for the last digit being 2 turned out to be unnecessary, but I

*think * that's a coincidence.

ETA: In reading through Mike's solution, I see the -4 makes sense if you declare x to be the number without the final 2. Sorry about that!

I wonder if we can prove we don't have to check the final digit, as both Mike and I discovered?