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 2007-06-23, 20:27 #1 davar55     May 2004 New York City 102038 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 641 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 $n = 10^{k+1}$ then there are $10^{k-1}-1$ from the ones digit, $10^{k-1}-10$ from the tens digit, and so on, thus if $n = 10^{k+1}$, then we have accumulated $\sum_{i=0}^{k-1} 10^{k-1} - 10^i$ 0s, this equals $k*10^{k-1} - \sum_{i=0}^{k-1} 10^i = k*10^{k-1} - \frac{10^k-1}{9}$ 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 3×1,409 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 3×1,409 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

24×33 Posts

Quote:
 Originally Posted by Orgasmic Troll if $n = 10^{k+1}$ then there are...
don't you mean n=10^k-1 ?

2007-07-10, 00:27   #6
m_f_h

Feb 2007

1101100002 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[48]:0,"
",($a=count_chars(join(range(0,999))))?$a[48]:0,"
",($a=count_chars(join(range(0,9999))))?$a[48]: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[1] + m[2]. Count 0's in m[1] and multiply this by m[2]+1: This
* is the number of 0's in { 10 m[1],..., n } minus 1 (trailing 0 of 10 m[1]).
*/
n=m; while( n=divrem(n[1],10), cnt += !n[2] ); cnt *= (m[2] + 1);
/* now add the number of 0's occuring in last position in {0,...,10 m[1]}
* (this is equal to 1+m[1]) plus 10 * the number of 0's in { 1,...,m[1]-1 }.
*/
cnt+1+m[1]+10*(cnt0( m[1]-1 )-1)
}

 2007-07-10, 20:32 #8 petrw1 1976 Toyota Corona years forever!     "Wayne" Nov 2006 Saskatchewan, Canada 10EF16 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 10010001111012 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 3·5·172 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.

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