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Topic: cotpi 63 - Same leading digit
Replies: 1   Last Post: Oct 21, 2012 2:34 AM

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James Waldby

Posts: 353
Registered: 1/27/11
Re: cotpi 63 - Same leading digit
Posted: Oct 21, 2012 2:34 AM
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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 brute-force 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 = c-b
~ .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 near-zero or
near-one 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 base-10 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 near-multiple of 1/2. Then l(3) < 1/2 < l(4) forces the
leading digit of 2^q, 5^q to be a 3.

--
jiw



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