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- Xref: sparky sci.math:14490 rec.puzzles:7084
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- From: baez@guitar.ucr.edu (john baez)
- Newsgroups: sci.math,rec.puzzles
- Subject: Math puzzles - answer
- Message-ID: <23678@galaxy.ucr.edu>
- Date: 5 Nov 92 20:30:37 GMT
- Sender: news@galaxy.ucr.edu
- Followup-To: sci.math
- Organization: University of California, Riverside
- Lines: 47
- Nntp-Posting-Host: guitar.ucr.edu
-
- Robert Southworth, bog@ugcs.caltech.edu, asked me to post this for him:
-
- John Baez (jbaez@riesz.mit.edu) posted two math puzzles
- <1992Oct28.170601.19561@galois.mit.edu>, to which he later gave
- partial answers. He writes:
-
- > The second question concerned the sum from 1 to infinity of o(2^n)/2^n.
- > Here o(2^n) is the number of odd digits in the decimal representation of
- > 2^n. The sum is surprisingly simple - it's 1/9.
-
- (1) Clearly the kth digit (the multiplier of 10^k in the decimal
- representation) of 2^n is odd iff
-
- [2^n/10^k] is odd,
-
- where [] denotes the greatest integer function.
-
- (2) Similarly, the (-n)th bit (the multiplier of 2^(-n), in the binary
- representation) of 10^(-k) is odd (i.e. equal to 1) iff
-
- [10^(-k)/2^(-n)] = [2^n/10^k] is odd.
-
-
- Let d_nk be congruent to [2^n/10^k] mod 2, d_nk = 0 or 1. Then by (1),
-
- o(2^n) = sum_k d_nk,
-
- and by (2),
-
- 10^(-k) = sum_n (d_nk * 2^(-n)).
-
- The result follows by reversing the order of summation, as follows:
-
- sum_n o(2^n)/2^n = sum_n (sum_k d_nk)/2^n
- = sum_k (sum_n d_nk/2^n)
- = sum_k 10^(-k)
- = 1/9.
-
- (Note that 10^0 is not included in the final sum over k, because the
- 0th digit of 2^n is even for all n>=1.)
-
- This problem is easy to generalize. The particular values 2 and 10 are
- not important; their only crucial property is that 10 is divisible by 2.
-
-
- Robert Southworth bog@ugcs.caltech.edu
-
-