Leftmost Digits of 2^nDate: 11/01/2002 at 16:06:41 From: Claudio Buffara Subject: Leftmost digits of 2^n Hi, Here is my problem: Prove that there is a power of 2 whose decimal representation starts with the digits 1999. A hint was given: log of 10 in the base 2 is irrational. The way I see it, I must prove there exist positive integers m, n, and A such that: 2^m = 1999 * 10^n + A, where 1 <= A <= 10^n - 1. I can't see any obvious way to use the hint. I did some algebraic manipulations, such as: A = 2^m - 1999*10^n = 2^n*(2^(m-n) - 1999*5^n) and deduced that: 2^n || A ==> A = 2^n * b, where b is odd. Also, 1 <= A <= 10^n - 1 ==> 1 <= b <= 5^n - 1. And that was as far as I got. I'd really appreciate some help from you. Thanks and regards, Claudio Buffara. Date: 11/01/2002 at 17:03:59 From: Doctor Schwa Subject: Re: Leftmost digits of 2^n Hi Claudio, To use the hint, let's rewrite your statement about 2^m and A instead as proving that there's some m and n so that 1999 * 10^n < 2^m < 2000 * 10^n Taking log base 10 of both sides, log(1999) + n < m log 2 < log(2000) + n and you know log 2 is irrational. Can you use that to prove that some multiple of log 2 is between log(1999) and log(2000), mod 1? (mod 1 means taking the fractional part; the n can take care of the integer part.) - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 11/04/2002 at 08:45:09 From: Claudio Buffara Subject: Leftmost digits of 2^n Got it! Your inequalities: n + log(1999) < m*log(2) < n + log(2000) can be rewritten as: log(1999) < -n + m*log(2) < log(2000). However, given that log(2) is irrational, it is true that for any epsilon > 0, there are positive integers a and b such that: 0 < | -a + b*log(2) | < epsilon - a result that can be proved by means of the pigeonhole principle (in fact, I never saw a proof of that which doesn't use the PHP - I'd be very interested to see if one exists). That, in turn, implies that the numbers of the form '-p + q*log(2)' with p & q positive integers are dense in R (given any interval (u,v), set epsilon = (v-u)/2 in the above inequality and take a and b such that 0 < | -a + b*log(2) | < (v-u)/2. Then, by the Archimedean property of R, some integer multiple of -a + b*log(2) must fall inside (u,v) ). In particular, there are positive integers m and n such that: log(1999) < -n + m*log(2) < log(2000) ==> 1999 < 10^(-n) * 2^m < 2000 ==> 1999 * 10^n < 2^m < 2000 * 10^n ==> the first four digits of 2^n are 1999..... Thanks a lot for your help. Best regards, Claudio Buffara. Date: 11/04/2002 at 09:55:06 From: Claudio Buffara Subject: Leftmost digits of 2^n I think the problem has a sweeping generalization: Let P be a positive integer that is not a power of 10 (so that log(P) is irrational) and let A1,A2,...,Ar be any sequence of decimal digits. Then there is a power of P whose decimal representation starts with the digits A1...Ar. Let A = A1*10^(r-1) + A2*10^(r-2) + ... + Ar. We must show that it is always possible to find positive integers m and n such that: A*10^n < P^m < (A+1)*10^n. But due to the irrationality of log(P), the proof is identical to the one for the power of 2. Best regards, Claudio Buffara. Date: 11/04/2002 at 13:18:07 From: Doctor Schwa Subject: Re: Leftmost digits of 2^n Exactly so. You can make the leftmost digits be any string you want, as long as log(P) is irrational. It's quite a different story if you want to study the rightmost digits. That's a number theory problem, and quite a fun one. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/