 mersenneforum.org Find the Value
 Register FAQ Search Today's Posts Mark Forums Read  2007-06-23, 20:27 #1 davar55   May 2004 New York City 108B16 Posts Find the Value Consider the set of integers from 0 to n represented in decimal with no lead zeros. Find the value of n > 1 such that the total number of zeros in the representations equals n.   2007-06-26, 22:44 #2 Orgasmic Troll Cranksta Rap Ayatollah   Jul 2003 64110 Posts so here's what I do know.. if n = 999, then we have 99 0s from the ones digit and 90 0s from the tens digit if n = 9999, we have 999 0s from the ones digit, 990 0s from the tens digit and 900 0s from the hundreds digit if then there are from the ones digit, from the tens digit, and so on, thus if , then we have accumulated 0s, this equals now, somewhere in between 10^11-1 and 10^12-1, we have more 0s than n, in fact, we have 88,888,888,889 more 0s, so, assuming that the difference is monotone increasing (I believe it is, although I haven't checked formally) it has to happen for some 11 digit number (or it doesn't happen at all)   2007-06-27, 17:51 #3 davar55   May 2004 New York City 108B16 Posts Please double-check that you don't mean some 12-digit number, since the transition point is > 10^11 - 1. And don't forget that 0 is part of the set, so you might need to add 1 to the number of zeros.   2007-07-09, 17:49 #4 davar55   May 2004 New York City 108B16 Posts An update: Upon analysis, the number of zeros in the representations of integers from 0 up to 99,999,999,999 is 98,888,888,890. Iterating from that point gives a match at n = 100,559,404,365. [Would someone double-check this value?]   2007-07-10, 00:11   #5
m_f_h

Feb 2007

1B016 Posts Quote:
 Originally Posted by Orgasmic Troll if then there are...
don't you mean n=10^k-1 ?   2007-07-10, 00:27   #6
m_f_h

Feb 2007

24×33 Posts Quote:
 Originally Posted by Orgasmic Troll if n = 999, then we have 99 0s from the ones digit and 90 0s from the tens digit if n = 9999, we have 999 0s from the ones digit, 990 0s from the tens digit and 900 0s from the hundreds digit
Code:
gp > ^Z
Suspended
(mfh@lx08 ~/NumberTheory/PARI) php
<?=($a=count_chars(join(range(0,99))))?$a:0,"
",($a=count_chars(join(range(0,999))))?$a:0,"
",($a=count_chars(join(range(0,9999))))?$a:0,"
";^D
10
190
2890
(mfh@lx08 ~/NumberTheory/PARI) fg
gp
gp > calc0mfh(k)=k*10^(k-1)-(10^k-1)/9+1
time = 0 ms.
gp > calc0mfh(3)
time = 0 ms.
%216 = 190
gp > calc0mfh(4)
time = 0 ms.
%217 = 2890
gp >
thus, the correct formula is:
number of '0's in {0,1,...,10^(k+1)-1} = k*10^(k-1)-(10^k-1)/9+1

I confirm your value for k=11 which is correct, i.e. seems to include the +1 already.
If your "iterations"(?) are ok, the result should be correct.

Last fiddled with by m_f_h on 2007-07-10 at 01:26   2007-07-10, 04:08   #7
m_f_h

Feb 2007

24×33 Posts Quote:
 Originally Posted by davar55 [Would someone double-check this value?]
it seems ok - I believe the following code is correct:
Code:
cnt0(n) = { local( cnt=0, m=divrem(n,10)); if( n<10, return(1));
/* Let n = 10 m + m. Count 0's in m and multiply this by m+1: This
* is the number of 0's in { 10 m,..., n } minus 1 (trailing 0 of 10 m).
*/
n=m; while( n=divrem(n,10), cnt += !n ); cnt *= (m + 1);
/* now add the number of 0's occuring in last position in {0,...,10 m}
* (this is equal to 1+m) plus 10 * the number of 0's in { 1,...,m-1 }.
*/
cnt+1+m+10*(cnt0( m-1 )-1)
}   2007-07-10, 20:32 #8 petrw1 1976 Toyota Corona years forever!   "Wayne" Nov 2006 Saskatchewan, Canada 2×2,417 Posts Without a lot of deep thought I surmised that there would be an equal distribution of digits in counting up from 1 ... but I discovered that, while that is the case it is not until 'n' gets VERY LARGE For example: we get the correct number of digits of 1 at only n=199,981 and many more times as n increases. However, digits 2 - 9 match much later ... 2 at 242,463,827; 3 at 371,599,983 (I haven't double checked these, though)   2007-07-10, 22:13   #9
m_f_h

Feb 2007

24×33 Posts Quote:
 Originally Posted by petrw1 For example: we get the correct number of digits of 1 at only n=199,981 and many more times as n increases. However, digits 2 - 9 match much later ... 2 at 242,463,827; 3 at 371,599,983 (I haven't double checked these, though)
I don't know if this is related but I once read that 1 is also the most common digit in any "random" collection of numbers (e.g. table of physical constants etc)
which has to do with the fact that the range [1,2) is "relatively" larger than [2,3) etc. (I don't remember well how exactly the argument was going - maybe w.r.t. scientific notation (x=m*10^e) or logarithmic scale or....)   2007-07-11, 04:12 #10 axn   Jun 2003 517910 Posts Perhaps you're thinking about Benford's Law   2007-07-11, 17:56 #11 petrw1 1976 Toyota Corona years forever!   "Wayne" Nov 2006 Saskatchewan, Canada 2·2,417 Posts When I wrote a simple program to simply go through all values of n starting from 1 and counting the occurence of each digit, whenever I got to a n of 99...99 the total number of each digit from 1 to 9 was the same with 0 consistently trailing ... but gaining. n=9: 0=0; 1-9=1 n=99: 0=9; 1-9=20 n=999: 0=189; 1-9=300 n=9999: 0=2889; 1-9=4000 n=99999: 0=38889; 1-9=50000 .... and the pattern continues. I think I just described in layman terms what the true mathemiticians described earlier in formulae.   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Stargate38 Programming 18 2015-07-10 06:08 davar55 Math 2 2010-02-19 16:54 davar55 Puzzles 7 2009-07-02 19:46 maheshexp Miscellaneous Math 29 2004-08-30 15:59 maheshexp Software 2 2004-05-08 03:16

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