| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 968.1 | I worked hard at it, but all for naught | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Fri Nov 04 1988 20:25 | 13 |
| Spoiler follows, so "next unseen" because you really do want
to work on this one yourself.
By the way, after you have solved it (or peeked, no more
warnings) go back and look at the title of this topic.
The only thing I noticed about the series is that it leaves
out the numbers whose factorials in decimal have embedded
zeroes. So the numbers in the sequence have all of the
non-zero digits before any/all of the zero digits in their
factorial as written in base ten.
Dan
|
| 968.2 | and then there is... | AKQJ10::YARBROUGH | I prefer Pi | Mon Nov 21 1988 16:30 | 10 |
| Ah, but what is the next number in the sequence?
Author's intended solution follows:
40! = 815915283247897734345611269596115894272000000000
I'm pretty sure that's the last factorial with only trailing zeroes.
Lynn
|
| 968.3 | | ELIS::GARSON | V+F = E+2 | Tue Jan 29 1991 09:19 | 7 |
| re .2
> 40! = 815915283247897734345611269596115894272000000000
>
>I'm pretty sure that's the last factorial with only trailing zeroes.
Can you prove this? What was the evidence for the conjecture?
|
| 968.4 | | TRACE::GILBERT | Ownership Obligates | Tue Jan 29 1991 13:06 | 10 |
| > Can you prove this? What was the evidence for the conjecture?
The conjecture is true thru 200!, which is a 375-digit number that has 49
trailing zeroes.
Suppose that the 326 digits before the trailing zeroes in 200! were randomly
chosen. What is the likelyhood that none of these is a zero? About 0.9**326,
or 1.21x10**-15, which is pretty unlikely -- about one in a *quadrillion*.
For larger factorials, the `probability' of no zeroes quickly becomes more
remote. Hence the conjecture.
|
| 968.5 | | ELIS::GARSON | V+F = E+2 | Tue Jan 29 1991 14:42 | 20 |
| >The conjecture is true thru 200!, which is a 375-digit number that has 49
>trailing zeroes.
Assuming my program is correct, I verified it out to 2000! but that
gives only empirical evidence.
>Suppose that the 326 digits before the trailing zeroes in 200! were randomly
>chosen. What is the likelyhood that none of these is a zero? About 0.9**326,
>or 1.21x10**-15, which is pretty unlikely -- about one in a *quadrillion*.
>For larger factorials, the `probability' of no zeroes quickly becomes more
>remote. Hence the conjecture.
Agreed. I haven't investigated the assumption that the digits are
uniformly and independently distributed. I was just thinking it would
be nice if someone could prove the conjecture. I looked at it briefly
and couldn't see any promising lines of attack.
An unrelated but illustrative example of where the above assumption
fails is the distribution of the leading digit of powers of 2 (in base
10). [An interesting problem in its own right.]
|
| 968.6 | Perhaps this is easier | ELIS::GARSON | V+F = E+2 | Mon Feb 04 1991 09:15 | 12 |
| re .*
Perhaps the same problem but with the factorials expressed in binary is
easier. The "no embedded zero" condition then becomes equivalent to
finding (integer) solutions for
m n
(2 - 1).2 = k!
which look like being k = 1,2,3,4,5 (m and n as appropriate).
m
Any takers? I'm sure a lot is known about factorising 2 - 1.
|
| 968.7 | let me try to summarize what we're asking about zeroes in factorials | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Mon Feb 04 1991 20:41 | 7 |
|
Are we asking:
Are there any integers above 40 whose factorial contains NO
embedded zeroes ?
|
| 968.8 | Yes | ELIS::GARSON | V+F = E+2 | Thu Feb 07 1991 09:10 | 6 |
| re .-1
The base noter conjectured that in base 10 representation there are no
such integers. This seemed too difficult so I am conjecturing that in
base 2 there are no integers above 5 whose factorial contains NO embedded
zeros.
|