For any k-digit number M there exist a number n such that the string of first k digits of 2^n equals M .

Prove it.

Well, remarkable as this is, it is not unexpected. In fact, it is exactly what I would expect based on my trusty slide rule.

On my slide rule, all numbers whose significant leading digits are M are on the same little stretch of tempered metal.

Start plotting all powers of 2 on the slide rule. The first one goes at roughly .301 (ie, the log 2), the 2nd at .602, the 3rd at .903, the 4th at .204 (the fractional part of 4*log2), the nth at n*log2 mod 1.

Because log 2 is irrational, there will eventually be many points in any and every selected little stretch on the slide rule. Thus, the proposition is proved.

Further, this is also true for 3^n, 4^n, 17^n, 3000^n, and any number that is not a perfect power of 10.

*Edited on ***January 30, 2011, 11:26 am**

*Edited on ***January 30, 2011, 2:17 pm**