Prove the following theorem:
For any string of decimal digits, representing number M there is an integer n such that the number N=2^n begins with the string M.
Find a value of N for M=2023?
The theorem was previously posted on this site in 2011, with the title "Start as you wish".
Here it is.
I am repeating my proof:
"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."
Old-fogey logic
