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Re: cotpi 63  Same leading digit
Posted:
Oct 21, 2012 2:34 AM


On Sat, 20 Oct 2012 11:07:38 0700, Barry Schwarz wrote: > On Sat, 20 Oct 2012 18:37:52 +0530, cotpi <puzzles@cotpi.com> wrote: >>The powers 2^n and 5^n start with the same digit >>where n is a positive integer. What are the possible >>values for this digit? > > The leading digit of a value x expressed in decimal notation can be > found with LD(x) = floor(x/(10^floor(log(x)))) > where log is the common (base 10) logarithm function. > > The desired values of n are those which satisfy > floor((2^n) / (10^floor(log(2^n)))) = > floor((5^n) / (10^floor(log(5^n)))) > > My limited bruteforce testing shows 5, 15, 78, 88, 98, 108, 118, 181, > 191, 201, and 211 are solutions. Oddly enough, the leading digit is 3 > in all these cases. > > If there is a significance to the digit 3 or that the solutions > cluster in groups of four separated by 10, I'm not smart enough to > discern it.
The separation by 10 (as in q = 78, 88, 98, 108, 118) arises because 3/10 and 7/10 appear among convergents of continued fractions for l(2) and l(5), (with l() denoting base 10 logarithms) or equivalently that 2^10 and 5^10 are close to powers of 10; that is, 1024 ~ 10^3 and 9765625 ~ 10^7. The reason for leading digit being 3 is explained in last paragraph below.
Writing b for l(2) = log(2)/log(10) and c for l(5), and x = cb ~ .397940008672..., the continued fraction for x is [0;2,1,1,18,1,4,...] which has convergents equal to 1/2, 1/3, 2/5, 37/93, 39/98, 193/485, 425/1068, 5293/13301, 5718/14369, 22447/56408, 28165/70777, ..., and you will note numbers 5 and 98 in the denominators of those convergents.
Each convergent p/q of a CF for value x is a best rational approximation to x; ie is closest to x among all reduced fractions with denominators no larger than q. When p/q ~ x we have q*x ~ p. As p is an integer, q*x will be approximately an integer, ie will have a nearzero or nearone fractional part. For example, with x ~ .397940008672... as above, the values 5*x, 93*x, 98*x, 485*x, and 1068*x are equal to about 1.989700, 37.00842, 38.99812, 193.0009, and 424.999929, respectively.
When q*x is nearly an integer p, then leading digits of 2^q and 5^q are more likely to be equal: Let l(m) denote the base10 log of m. From before, b=l(2), c=l(5). Now l(2^q) = q*l(2) = q*b, and l(5^q) = q*l(5) = q*c so l(5^q)  l(2^q) = q*c  q*b = q*x ~ p. Also, q*x = l(5^q /2^q) so q*x being nearly an integer means that 5^q and 2^q start with digit strings of similar value. (The same idea applies in other bases too. For example, convergents of l(7)l(3) include 4/11, 7/19, 39/106, 46/125. 3^11 and 7^11 begin with digit 1; 3^19 and 7^19 begin with digits 11; 3^106 and 7^106 begin with 3; 3^125 and 7^125 begin with 43.)
Example 1: q=1068, q*x~ 424.999929, 2^q~ .31625e321, 5^q~ .31620e746 Example 2: q=485, q*x ~ 193.0009, 2^q ~ .99895e145, 5^q~ .10010e339 Example 3: q=191, q*x ~ 76.0065, 2^q ~ .313855e57, 5^q~ .318618e133 Note, in example 3, 191 = 93 + 98.
The leading digit of 3 for the 2^q, 5^q case arises because 2*5 = 10 and l(2) + l(5) = l(10) = 1. If p/q is a convergent of x = l(5)l(2), there is a number f such that { f/(2q), (f+2p)/(2q) } are convergents of l(2) and l(5). When f is odd, q*x being nearly an integer forces f/(2q) to be an odd nearmultiple of 1/2. Then l(3) < 1/2 < l(4) forces the leading digit of 2^q, 5^q to be a 3.
 jiw



