Euler problem 57

It is possible to show that the square root of two can be expressed as an infinite continued fraction.

 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?

analyse:
If you examine the function: Log10(Numerator)/Log10(Denominator) 
you'll notice a pattern. It looks like that each 8th and 13th iteration 
yield to a fraction 2 (iteration 8,13,21,26,34,39,...)
So in 1000 iterations there are 2 positive interations every 13 iterations. 
=> answer: 2 * 1000 / 13 

Problem 57 =         153 elapsed time:    0 ms. Test Passed.