[Guest]

Can someone prove that there exist two positive integers, m and n, for which 2^m and 5ⁿ share same first ten (most significant) digits

Edited November 15, 2010 by Dhanannjay Deo

[EventHorizon]

I start with calculating log₂5.

Since 2^(a*log₂5) = 5^a, we need to find a positive integer that is very close to a*log₂5 for some a.

Since we want the first 10 significant digits to be the same, lets give a multiplicative error of 1.000000000000000000000001. This is much better than we need, and we could always choose a smaller error.

First take the log base 2 of the error, and that is the most that a*log₂5 can (additively) be away from a positive integer. So I'll look for a string of 0's longer than that in the decimal expansion of log₂5.

Assuming log₂5 is normal, you can find the string of 0's. If it is not normal, you can look for strings of any single repeated digit, and then multiply by 9, and the result should have the 0's. If it is not normal and none of these work, just look for a multiple of log₂5 that has a fractional part smaller than the desired error. (Just found a proof that you can always find an integer factor to get the fractional part as low as you want here, Yay for the pigeon-hole principle)

Once found, a is the power of 10 (or 9* the power of 10, or just the integer that must exist due to the proof) that makes the value a positive integer when ignoring values after the string of zeros. Let b be the value of floor(a*log₂5). Then (2^b - 5^a) is less than the chosen error, and should have the same first 10 significant digits (if not (the very unlikely case of a string of 9's and the value carrying up to the top ten significant digits) choose a smaller error, and repeat).

[EventHorizon]

Obviously, there are an infinite number of rational numbers between any two non-equal numbers.

So there are an infinite number of rationals between log₂5 and log₂5 + 10^-20 (or anything to get less than the multiplicative error needed).

Choose one, and let it be represented as a/b. Which should leave 2^a and 5^b with the same 10 most significant digits.

[Guest]

another option is to note that there are a finite number of ways you can get 10 digits, and since each power produces "random" numbers, they will eventually equal.

it's a bit like trying to find a finite sequence of pi within e. you'll do it though it will take a while.