http://elgordillo.blogspot.com/2004/09/another-derbyshire-maths-problem.html

Puzzle:

Consider the powers of two: 2, 4, 8, 16, 32, 64, 128… Their right-most digits follow a simple repetitive pattern: 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6… What about their left-most digits, though? Here are the first 40-odd [actually, it's the first 149 - Ed.]. (Read right to left, line by line. I’ve just jammed the digits together, leaving out the courtesy commas, to save space.)

24813612512481361251248136125

124813612512481371251249137125

124913712512491371361249137136

124913713612512481361251248136

125124813612512481361251248137…

There is a pattern there, but it keeps breaking down. When you get up into really big powers of 2, in fact, it breaks down altogether. For the ten powers of 2 from the 30th to the 39th, for instance, we have lead digits 1, 2, 4, 8, 1, 3, 6, 1, 2, 5. For the ten powers from the billionth to the 1,000,000,009th, by contrast, we have lead digits 4, 9, 1, 3, 7, 1, 2, 5, 1, 2. Every digit (except, of course, zero) shows up, though.

I am going to refer to this infinite sequence of digits as “the sequence.” Now I am going to ask the following two questions: In the first N digits of the sequence, how many occurrences of 3 shall I find? And how many occurrences of 4?

Call the first number U(N), the second V(N). The first few values of U(N) and V(N), for N from 1 to 15, look like this, as you can easily verify:

U(N) = 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2,…

V(N) = 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2,…

Show that for sufficiently large N, U(N) will always be bigger than V(N). Find the limit of the ratio U(N) to V(N) as N tends to infinity.