The four fours Problem
Breyfogle
I have been given the task to express each whole number within the range 1
thru 100
using only the digit '4' and the basic arithmatic operators +, -, *, /,
4th power,
factoral and sq root. Each expression must contain 4 fours. I have been
told that
there is a complete solution, but I have my doubts. Most odd numbers
above 30 are
particularly difficult.
As examples:
31 = 4! + (4!+4)/4
60 = (4 to power 4)/4 - 4
Does anyone recognize this problem as a known puzzle ? If so, I would
like to find
its history, solution and any reference links that discuss it.
Shawn Labounty
Breyfogle wrote:
>
> I have been given the task to express each whole number within the range 1
> thru 100
> using only the digit '4' and the basic arithmatic operators +, -, *, /,
> 4th power,
> factoral and sq root. Each expression must contain 4 fours. I have been
> told that
> there is a complete solution, but I have my doubts. Most odd numbers
> above 30 are
> particularly difficult.
>
Yes, it is a known probem. I'll take a look and see if I can verify
that all of the numbers 1-100 are attainable.
A couple "over 40's" for you:
44 - 4/4 = 43
44 + 4/4 = 45
--
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/:/_/:/\:\__\ /\ \ /\__\ /:/_/:/\:\__\ /:/_/:/\:|__|
\:\/:/ /:/ / \:\ \ /:/ / \:\/:/ \/__/ \:\/:/ /:/ /
\::/ /:/ / \:\ /:/ / \::/__/ \::/_/:/ /
\/_/:/ / \:\/:/ / \:\ \ \:\/:/ /
/:/ / \::/ / \:\__\ \::/ /
\/__/ \/__/ \/__/ \/__/
Shawn Labounty sl...@busy.enet.dec.com
Materials Analyst Excess Materials Group
Digital Equipment Corporation Marlboro, MA
Shawn Labounty
Breyfogle wrote:
>
> I have been given the task to express each whole number within the range 1
> thru 100
> using only the digit '4' and the basic arithmatic operators +, -, *, /,
> 4th power,
> factoral and sq root. Each expression must contain 4 fours. I have been
> told that
> there is a complete solution, but I have my doubts. Most odd numbers
> above 30 are
> particularly difficult.
>
>
> 31 = 4! + (4!+4)/4
>
> 60 = (4 to power 4)/4 - 4
>
> Does anyone recognize this problem as a known puzzle ? If so, I would
> like to find
> its history, solution and any reference links that discuss it.
I don't understand the explanation, but this is apparently correct.
Stop reading now if you don't want to see the answer.
Let log[k] = logarithm to base k.
Let sqrt[k] = sqrt operator replicated k times. Then
N = - log[sqrt(sqrt(4*4))] log[4] sqrt[N](4)
for all integer N.
(I hope I've typed that correctly.)
This rsult is not new; I think Martin Gardner published it several years
ago
in Mathematical Games/Scientific American.
Timothy J. Lipetz
As you stated it, the problem is not solvable. For example 55 can not
be done with the operators given. However, I always heard the problem
included the repeating decimal operator (a dot over the number). For
example, .4(with a dot over it) == .444444444... == 4/9. With that, all
100 numbers can be found.
A few years, ago I wrote a program to find all 4 4's (actually it was a
rather challenging exercize). I posted my results to this group at the
time.
By the way, 55 was...
( 4! / .4(with dot) ) + 4 / 4
== ( 24 / (4/9) ) + 1 == 55
Tim Lipetz
Eva R. Myers
In article <33938B...@netset.com>,
Timothy J. Lipetz <lip...@netset.com> wrote:
>As you stated it, the problem is not solvable. For example 55 can not
>be done with the operators given. However, I always heard the problem
>included the repeating decimal operator (a dot over the number). For
>example, .4(with a dot over it) == .444444444... == 4/9. With that, all
>100 numbers can be found.
>
>A few years, ago I wrote a program to find all 4 4's (actually it was a
>rather challenging exercize). I posted my results to this group at the
>time.
>
>By the way, 55 was...
>
>( 4! / .4(with dot) ) + 4 / 4
>
>== ( 24 / (4/9) ) + 1 == 55
A simpler solution is 55 = 44 / (.4 + .4), but this still doesn't meet
the criteria below since it uses decimal points.
How could you be sure that your program found all integer 4 4's within
the range you were interested in? (I put those qualifications on it
because you can make 4!, 4!!, 4!!!,... and do a similar thing with
square roots.) For example, I am convinced that you can't make 7 with
two 4s using only the operators mentioned in this post, ^ (to the
power, as in 4^4 = 256), and nth root of (as in ".4"th root of 4 =
32). But how would you prove that a suitable combination of
factorials and square roots won't give you 7? Do you have a general
proof that limits the number of factorials and square roots your
program needs to search through?
>> Breyfogle wrote:
>> >
>> > I have been given the task to express each whole number within the range 1
>> > thru 100
>> > using only the digit '4' and the basic arithmatic operators +, -, *, /,
>> > 4th power,
>> > factoral and sq root. Each expression must contain 4 fours. I have been
>> > told that
>> > there is a complete solution, but I have my doubts.
Eva Myers.
--
Eva Myers, Newnham College, Cambridge Email: erm...@cam.ac.uk
Ignorance and deception can't save anybody. *Knowing* saves them.
PGP public key obtainable from keyservers
Matthew Daly
erm...@phy.cam.ac.uk (Eva R. Myers), if that is your REAL name, said:
>
>A simpler solution is 55 = 44 / (.4 + .4), but this still doesn't meet
>the criteria below since it uses decimal points.
The criteria given isn't even very clear about 44. The original post
said that the _number_ 4 had to be used four times, not the _digit_
4.
>How could you be sure that your program found all integer 4 4's within
>the range you were interested in? (I put those qualifications on it
>because you can make 4!, 4!!, 4!!!,... and do a similar thing with
>square roots.) For example, I am convinced that you can't make 7 with
>two 4s using only the operators mentioned in this post, ^ (to the
>power, as in 4^4 = 256), and nth root of (as in ".4"th root of 4 =
>32). But how would you prove that a suitable combination of
>factorials and square roots won't give you 7? Do you have a general
>proof that limits the number of factorials and square roots your
>program needs to search through?
I don't think it's that hard to see. No factorial is a square (indeed,
no factorial is a perfect power of _any_ degree). Similarly, I
would argue that you can only take the factorial of positive integers,
so having square roots after factorials is quite limited. (Yes, I know
about the gamma function, but saying that ((1.5)!)^2 = pi just because
x! = gamma(x-1) for all positive integers is an abuse of the notation,
IMHO.) So, the only natural numbets that you can make from one four
are v4 = 2, 4, 4!=24, 4!!, and so on. (Notice that 2! = 2, so you don't
gain anything there.)
But there is no way to turn 4!! = 24! into a "tractable" number that can
be used to form numbers in the 1-100 range. Well, okay, 4!!/(4!-4/4)! = 24,
I suppose, but you can see the limited application. :-)
ObPuzzle: The problem asked that you use _exactly_ four 4's and the
operations of addition, subtraction, multiplication, division, factorial,
square root, and raising to the fourth power to form all numbers from 1-100.
Show that it is sufficient to show equations for these numbers using
_up to_ four 4's.
ObPuzzle2: 24 = 4!!/(4!-4/4)! is just about the ugliest derivation I've
ever seen. I managed to come up with an even uglier derivation for 30.
Can you find it? (Solution after a space break.)
SPOILER FOR #2:
v is square root.
30 = (v(4! + 4x4))! / 4! = (v36)!/4! = 6!/4!
-Matthew
--
Matthew Daly I feel that if a person has problems communicating
mwd...@kodak.com the very least he can do is to shut up - Tom Lehrer
My opinions are not necessarily those of my employer, of course.
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Shawn Labounty
Ian Lynagh wrote:
>
> Once upon a time, in the land of rec.puzzles, Shawn Labounty eloquently
> composed:
> >Let log[k] = logarithm to base k.
>
>
> Where does log appear in:
> only the digit '4' and the basic arithmatic operators
> +, -, *, /, 4th power, factoral and sq root.
>
> TTYL
> Ian
> --
> Ian Lynagh - i...@lynagh.demon.co.uk
> http://www.sn.no/~balchen/igloo/
>
> I loathe people who keep dogs. They are cowards who haven't got the guts
> to bite people themselves.
Well, I knew it looked familiar so I went looking for an answer, even
though it might not fit the wording of the puzzle exactly.
The solutions that I found that DID fit the wording of the answer
covered all but five of the numbers, if I remember correctly.
--
Phil Gibbs
In article <5n1bd7$k2p$1...@kodak.rdcs.Kodak.COM>, Matthew Daly
<da...@PPD.Kodak.COM> writes
>ObPuzzle2: 24 = 4!!/(4!-4/4)! is just about the ugliest derivation I've
>ever seen. I managed to come up with an even uglier derivation for 30.
>Can you find it? (Solution after a space break.)
Not quite sure what you mean by "ugliest" but
how about ((sqrt(sqrt(4))^(4!)-4)/(sqrt(4)!!!!!!!!!!!!!!!!!!!!!!!!!) ?
I'm sure this could be improved upon using recurring decimals.
ObPuzzle: which numbers less than a hundred can be done with
only two or three fours (allowing decimal points and recurring
decimals)?
Phil Gibbs
http://www.weburbia.com/ http://www.weburbia.demon.co.uk/
"When all you've got is a hammer, everything looks like a nail"
Tim Firman
"Timothy J. Lipetz" <lip...@netset.com> wrote:
>As you stated it, the problem is not solvable. For example 55 can not
>be done with the operators given.
author omitted, but which you included, and which were allowed in the presumed
original 1881 formulation of this problem.
Here's 55 without your .4444444 operator:
44/(.4*sqrt(4)) = 55 (Not .44444... ; just 0.4)
This problem is solvable (without your operator) up to at least 200, and is
believed, but not proven, to be solvable for any integer if given sufficient
parenthesis, square roots, and factorials. (Not proven as of 1964)
It was solved to 112 before the turn of the century.
See _536 Curious Puzzles and Problems_, by Dudeney, the 'two fours' problem.
Martin Gardener wrote a column on it sometime in the 1960's, in Scientific
American, IIRC, and edited _536_ as well.
There are references within _536_ to a number of publications about it.
Tim
Phil Gibbs
In article <5n1kvk$16...@news.doit.wisc.edu>, Tim Firman <firman@?.?>
writes
>This problem is solvable (without your operator) up to at least 200, and is
>believed, but not proven, to be solvable for any integer if given sufficient
>parenthesis, square roots, and factorials. (Not proven as of 1964)
>It was solved to 112 before the turn of the century.
Don't you need something like the floor function to go beyond 112?
Tim Firman
da...@PPD.Kodak.COM (Matthew Daly) wrote:
>ObPuzzle2: 24 = 4!!/(4!-4/4)! is just about the ugliest derivation I've
>ever seen. I managed to come up with an even uglier derivation for 30.
>Can you find it? (Solution after a space break.)
>
>
>
>v is square root.
>
>30 = (v(4! + 4x4))! / 4! = (v36)!/4! = 6!/4!
With three fours(i realize you were looking for ugly, but I can't resist...):
30= (4!/4)!/4!
or 4!+4!/4
Well, 64 with two fours is:
vv((v4)^4!) = vv(2^24) = v(2^12) = 2^6 = 64
which has more mess per four, anyway...
(lifted from _536_)
an uglier but less useful one:
(vvvv(4!^(4!)))/(4^4) = 54
Tim
Jim Cross
The solution to 113 would make my day. Please send a note, if you have
it, anyone:)
Jim Cross
Phil Gibbs
In article <3395DD...@cinternet.net>, Jim Cross
<vact...@cinternet.net> writes
>The solution to 113 would make my day. Please send a note, if you have
>it, anyone:)
>
If you allow the floor function (square brackets) and recurring decimals
(brackets round repeated digits) it can be done using only three fours.
spoiler
.
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113 = [4^4*.(4)]
Courtenay Footman
In article <VmPQSHAT...@weburbia.demon.co.uk>,
Phil Gibbs <philip...@pobox.com> wrote:
>In article <3395DD...@cinternet.net>, Jim Cross
><vact...@cinternet.net> writes
>>The solution to 113 would make my day. Please send a note, if you have
>>it, anyone:)
>>
>
>If you allow the floor function (square brackets) and recurring decimals
>(brackets round repeated digits) it can be done using only three fours.
>
common than factorial, it is not analytic, and if one has floor,
square root, and factorial, it is far too easy to get anything.
--
-------------------------------------------------------------------------------
Courtenay Footman I have again gotten back on the net, and
c...@lightlink.com again I will never get anything done.
Matthew Daly
Phil Gibbs <philip...@pobox.com>, if that is your REAL name, said:
>In article <3395DD...@cinternet.net>, Jim Cross
><vact...@cinternet.net> writes
>>The solution to 113 would make my day. Please send a note, if you have
>>it, anyone:)
>>
>
>If you allow the floor function (square brackets) and recurring decimals
>(brackets round repeated digits) it can be done using only three fours.
If you can make a number with three fours, then you can make it with four.
> 113 = [4^4*.(4)]
or, 113 = vv[4^4*(.(4)]^4
Wei-Hwa Huang
Tim Firman <firman> writes:
>See _536 Curious Puzzles and Problems_, by Dudeney, the 'two fours' problem.
>Martin Gardener wrote a column on it sometime in the 1960's, in Scientific
>American, IIRC, and edited _536_ as well.
>There are references within _536_ to a number of publications about it.
Including the interesting article written by the computer scientist
Donald Knuth (author of Metafont and "The Art of Computer Programming")
when he was still at Caltech titled "Expressing Numbers With Only
One 4", published in American Mathematical Monthly.
--
Wei-Hwa Huang, whu...@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Question everything. Learn something. Answer nothing. -- Engineer's Motto
Imp...@aol.com
Hi. The answers to 33 and 39 would REALLY help me out. Thanks.
-------------------==== Posted via Deja News ====-----------------------
http://www.dejanews.com/ Search, Read, Post to Usenet
Matthew Daly
Imp...@aol.com, if that is your REAL name, said:
>Hi. The answers to 33 and 39 would REALLY help me out. Thanks.
As usual for this puzzle, you really need to specify which operations
are and aren't allowed for the puzzle to make sense.
But, let:
v4 = square root of 4
.4. = .4444... = 4/9
Therefore, 4/.4. = 9, a very useful derivation. Just as useful is
v(4/.4.) = 3.
So,
33 = 24 + 9 = 4! + 4/.4.
39 = 48 - 9 = 4! + 4! - 4/.4.
(Yes, the expression for 33 has only three 4's, but given any expression
X, vv(X^4) is an expression for the same number that contains an extra
4.)
Tim Firman
da...@PPD.Kodak.COM (Matthew Daly) wrote:
>Imp...@aol.com, if that is your REAL name, said:
>>Hi. The answers to 33 and 39 would REALLY help me out. Thanks.
>39 = 48 - 9 = 4! + 4! - 4/.4.
Or, without using the .4. operator:
33= (vv((v4)^(4!))+v4)/v4
39= 4!+4!/(4*.4)
Tim
Sten Thaning
Tim Firman <fir...@dabulls.chem.wisc.DELETETHISedu> wrote:
>>33 = 24 + 9 = 4! + 4/.4.
>>39 = 48 - 9 = 4! + 4! - 4/.4.
>
>Or, without using the .4. operator:
>
>33= (vv((v4)^(4!))+v4)/v4
>39= 4!+4!/(4*.4)
Hm. Do I do something terribly wrong?
4! + 4! / (4*.4) = (24 + 24) / 1.6 = 48/1.6 = 30, not 39.
The 33 one seems to be right, though.
- Sten Thaning
M Ivon M
Sten.Tha...@student.uu.se (Sten Thaning) wrote:
>Tim Firman <fir...@dabulls.chem.wisc.DELETETHISedu> wrote:
>>>33 = 24 + 9 = 4! + 4/.4.
>>>39 = 48 - 9 = 4! + 4! - 4/.4.
>>
>>Or, without using the .4. operator:
>>
>>33= (vv((v4)^(4!))+v4)/v4
>>39= 4!+4!/(4*.4)
>Hm. Do I do something terribly wrong?
Yep :)
>4! + 4! / (4*.4) = (24 + 24) / 1.6 = 48/1.6 = 30, not 39.
You put ( ) around 24 + 24.
Remove them and you get
24 + 24/1.6 = 24 + 15 = 39
>The 33 one seems to be right, though.
> - Sten Thaning
M Terribly Wrong M
Christ van Willegen
Sten Thaning wrote:
>
> Tim Firman <fir...@dabulls.chem.wisc.DELETETHISedu> wrote:
> >39= 4!+4!/(4*.4)
>
> Hm. Do I do something terribly wrong?
>
>
You read it the wrong way. As written, it means:
4! + (4!/(4*.4)) = 24 + (24/1.6) = 24+15 = 39
It is correct as stated.
Christ van Willegen
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Sten Thaning
abi...@fnx.com (Abigail) wrote:
>++ Hm. Do I do something terribly wrong?
>++
>++ 4! + 4! / (4*.4) = (24 + 24) / 1.6 = 48/1.6 = 30, not 39.
>Since when has + higher precedence than / ?
>4! + 4! / (4 * .4) = 24 + 24 / 1.6 = 24 + 15 = 39.
<blushing>
OK. I did something terribly wrong. Sorry. I will now go and put my
head in a bucket of water.
- Sten Thaning
Martin DeMello
>You read it the wrong way. As written, it means:
>
>4! + (4!/(4*.4)) = 24 + (24/1.6) = 24+15 = 39
>
>It is correct as stated.
i prefer
44-4-gamma(sqrt(4))
where gamma(n) = (n-1)! for integer n.
basically i dislike solutions involving .4 or .4(bar) since they make
the (unwarranted IMHO) assumption of base 10 arithmetic.
Martin
Phil Gibbs
In article <33a1c73e...@news.emirates.net.ae>, Martin DeMello
<vi...@emirates.net.ae> writes
>i prefer
> 44-4-gamma(sqrt(4))
>
>where gamma(n) = (n-1)! for integer n.
>basically i dislike solutions involving .4 or .4(bar) since they make
>the (unwarranted IMHO) assumption of base 10 arithmetic.
and using 44 doesn't?
Abigail
Martin DeMello (vi...@emirates.net.ae) wrote on 1383 September 1993 in
<URL: news:33a1c73e...@news.emirates.net.ae>:
++
++ i prefer
++ 44-4-gamma(sqrt(4))
++
++ where gamma(n) = (n-1)! for integer n.
++ basically i dislike solutions involving .4 or .4(bar) since they make
++ the (unwarranted IMHO) assumption of base 10 arithmetic.
44 - 4 == 40. gamma(sqrt(4)) == 1. But 40 - 1 being equal to 39
makes the assumption of base 10 arithmetic.
Abigail
Noam Elkies
In article <33a1c73e...@news.emirates.net.ae>,
>>4! + (4!/(4*.4)) = 24 + (24/1.6) = 24+15 = 39
>>It is correct as stated.
>i prefer
> 44-4-gamma(sqrt(4))
>where gamma(n) = (n-1)! for integer n.
>basically i dislike solutions involving .4 or .4(bar) since they make
>the (unwarranted IMHO) assumption of base 10 arithmetic.
Well 44 does also (as pointed out already); 44-4 can be replaced
sqrt(sqrt(sqrt(4^(4!)))) - 4! . But gamma() is not quite kosher
either.
--Noam D. Elkies (elk...@math.harvard:edu)
Dept. of Mathematics, Harvard University
Zeno Of Elea
Hello, I saw the old 4 4's problems being discussed once again, and I
thought I'd pop in with one I never could get.... I worked on these for
about 6 months one time... I got all of them 1-100 except for 89. If anyone
out there has the four-four of 89, not useing a repeated squaroot (to equal
1) or the .4(bar) (.44444...) Please let me know.... Thanks!!!
Thanks!!! :)
-----
Stephen M. Sebeny
sse...@neo.lrun.com
Martin DeMello
On 15 Jun 1997 13:45:07 GMT, elk...@ramanujan.harvard.edu (Noam
Elkies) wrote:
>In article <33a1c73e...@news.emirates.net.ae>,
>Martin DeMello <vi...@emirates.net.ae> wrote:
>>>You read it the wrong way. As written, it means:
>
>>>4! + (4!/(4*.4)) = 24 + (24/1.6) = 24+15 = 39
>
>>>It is correct as stated.
>
>>i prefer
>> 44-4-gamma(sqrt(4))
>
>>where gamma(n) = (n-1)! for integer n.
>>basically i dislike solutions involving .4 or .4(bar) since they make
>>the (unwarranted IMHO) assumption of base 10 arithmetic.
>
>Well 44 does also (as pointed out already); 44-4 can be replaced
>sqrt(sqrt(sqrt(4^(4!)))) - 4! . But gamma() is not quite kosher
>either.
You're right - i didn't notice the problem with 44 (removing plank
from eye). But why is gamma any less kosher than factorial? Also,
could someone tell me the best known soln. for 153?
Martin
Srikant Sridevan
ok. how about
89 = 65 + 24
89 = (4! + srqt4)/.4 + 4!
Its possible to do this 150 for sure.
--
Mails to s...@eos.ncsu.edu will bounce!
To email please replace the eos with apollo.psrc
Zeno Of Elea
In article <33A490...@eos.ncsu.edu>, Srikant Sridevan
<s...@eos.ncsu.edu> wrote:
> Zeno Of Elea wrote:
> >
> > Hello, I saw the old 4 4's problems being discussed once again, and I
> > thought I'd pop in with one I never could get.... I worked on these for
> > about 6 months one time... I got all of them 1-100 except for 89. If anyone
> > out there has the four-four of 89, not useing a repeated squaroot (to equal
> > 1) or the .4(bar) (.44444...) Please let me know.... Thanks!!!
OOPS!!! Sorry, I meant 87!!! If any of you have 87, PLEASE, PLEASE send it
to me... This would truely mean a lot to me... If there are any you don't
have ask me...
Fred W. Helenius
zenoo...@hotmail.com (Zeno Of Elea) wrote:
>I got all of them 1-100 except for 89. If anyone
>out there has the four-four of 89, not useing a repeated squaroot (to equal
>1) or the .4(bar) (.44444...) Please let me know.... Thanks!!!
89 = (4! + sqrt(4))/.4 + 4!
--
Fred W. Helenius <fr...@ix.netcom.com>
Matthew Daly
vi...@emirates.net.ae (Martin DeMello), if that is your REAL name, said:
>>You read it the wrong way. As written, it means:
>>
>>4! + (4!/(4*.4)) = 24 + (24/1.6) = 24+15 = 39
>>
>>It is correct as stated.
>
>i prefer
> 44-4-gamma(sqrt(4))
>
>where gamma(n) = (n-1)! for integer n.
>basically i dislike solutions involving .4 or .4(bar) since they make
>the (unwarranted IMHO) assumption of base 10 arithmetic.
Different strokes. I always thought that gamma was an ugly hack. You
get 1 and 6 with one symbol, which makes everything fall into place too
easily IMHO. Throw in that and the "triangle" symbol (for 1+2+...+n)
and you can probably get up into the thousands without breaking a sweat.
Oh well, at least it's better than adding the floor funciton, I guess.
And, as someone pointed out, 44 assumes base 10 arithmetic. If you
really wanted to get around it, you should say
39 = 4! + 4*4 - gamma(sqrt(4)) = 24 + 16 - 1
ObPuzzle: Give yourself gamma and triangle, but eliminate the "base-10"
operations. Can you form 9, 10, and 44 with one or two 4's? Would that
make the "baseless" system equally powerful?
Matthew Daly
zenoo...@hotmail.com (Zeno Of Elea), if that is your REAL name, said:
>
> Zeno Of Elea wrote:
> >
> > Hello, I saw the old 4 4's problems being discussed once again, and I
> > thought I'd pop in with one I never could get.... I worked on these for
> > out there has the four-four of 89, not useing a repeated squaroot (to equal
> > 1) or the .4(bar) (.44444...) Please let me know.... Thanks!!!
>
>to me... This would truely mean a lot to me... If there are any you don't
>have ask me...
I'm mildly surprised that you could get the rest of the numbers without
4/9. It's just such a handy way to get to the higher odd numbers. If
I might ask, how did you calculate 73 without it?
And there is a wide gulf of difference between using the repeated decimal
to represent 4/9 and somehow applying an infinite number of square root
operations to turn 4 into 1. The former is mathematically valid.
Martin DeMello
On 16 Jun 1997 18:25:25 GMT, da...@PPD.Kodak.COM (Matthew Daly) wrote:
>zenoo...@hotmail.com (Zeno Of Elea), if that is your REAL name, said:
>>
>> Zeno Of Elea wrote:
>>
>>OOPS!!! Sorry, I meant 87!!! If any of you have 87, PLEASE, PLEASE send it
>>to me... This would truely mean a lot to me... If there are any you don't
>>have ask me...
87 = 44*sqrt(4) - gamma(sqrt(4)) [Unfortunately with a 44]
>
>I'm mildly surprised that you could get the rest of the numbers without
>4/9. It's just such a handy way to get to the higher odd numbers. If
>I might ask, how did you calculate 73 without it?
73 = 4*4! - 4! + gamma(sqrt(4))
I reached 152 (lots of long, dull lectures :-) ) but using both 44 and
gamma(). Irritatingly, I feel sure that i've solved 153 sometime, but
can't for the life of me remember it.
>And there is a wide gulf of difference between using the repeated decimal
>to represent 4/9 and somehow applying an infinite number of square root
>operations to turn 4 into 1. The former is mathematically valid.
>
If you're willing to accept gamma(), gamma(sqrt(4)) is a handy way to
reach 1 using only one 4. It's also handy for gamma(4)=6.
Of course, there's always
n = -sqrt(4)*log_4(log_4(sqrt(sqrt(....ntimes...sqrt(4)))...)
Martin.
Seth Breidbart
In article <5o3lhl$bk6$1...@kodak.rdcs.Kodak.COM>,
Matthew Daly <da...@PPD.Kodak.COM> wrote:
>ObPuzzle: Give yourself gamma and triangle, but eliminate the "base-10"
>operations. Can you form 9, 10, and 44 with one or two 4's?
10 = triangle(4)
9 = triangle(4) - gamma(sqrt(4))
I don't see an easy way to get 44 with two 4's.
> Would that
>make the "baseless" system equally powerful?
Not necessarily; it still doesn't give 4444 with four 4's, or .4, etc.
Seth
Wei-Hwa Huang
da...@PPD.Kodak.COM (Matthew Daly) writes:
>ObPuzzle: Give yourself gamma and triangle, but eliminate the "base-10"
Yes.
Don Knuth, "Representing Numbers Using Only One 4." In
_Mathematics Magazine_, Vol 37, November/December 1964, pp. 308-10.
Shows how numbers up to 208 may be represented using only one
four, square roots, factorials and brackets [floor]. Conjectures
that all integers can be represented in this way. A classic
case of someone with way too much time.
ObPuzzle: What was the first publication to ever accept an article
written by Donald Knuth? The answer may surprise you.
--
Wei-Hwa Huang, whu...@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
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"Oh, that was when the sand was so hot we were each hopping on one foot."
Courtenay Footman
>>>You read it the wrong way. As written, it means:
>>>
>>>4! + (4!/(4*.4)) = 24 + (24/1.6) = 24+15 = 39
>>>
>>>It is correct as stated.
>>
>>i prefer
>> 44-4-gamma(sqrt(4))
>>
>>where gamma(n) = (n-1)! for integer n.
>>basically i dislike solutions involving .4 or .4(bar) since they make
>>the (unwarranted IMHO) assumption of base 10 arithmetic.
>
>Different strokes. I always thought that gamma was an ugly hack. You
>get 1 and 6 with one symbol, which makes everything fall into place too
>easily IMHO. Throw in that and the "triangle" symbol (for 1+2+...+n)
>and you can probably get up into the thousands without breaking a sweat.
>Oh well, at least it's better than adding the floor funciton, I guess.
>
seen it; it is certainly not a common operator.
Now one operator that I have seen (Gradshteyn uses it) is the bifactorial
operator, '!!', where n!! = product n(n-2)(n-4)... down to 2 or 1.
Thus 4!! = 8 and (Gamma(4))!! = 48.
Matthew Daly
c...@light.lightlink.com (Courtenay Footman), if that is your REAL name, said:
>>
>>Different strokes. I always thought that gamma was an ugly hack. You
>>get 1 and 6 with one symbol, which makes everything fall into place too
>>easily IMHO. Throw in that and the "triangle" symbol (for 1+2+...+n)
>>and you can probably get up into the thousands without breaking a sweat.
>>Oh well, at least it's better than adding the floor funciton, I guess.
>>
>What is the symbol for the triangle operator? I am not sure I have ever
>seen it; it is certainly not a common operator.
It's just a triangle. So you would say <triangle> 4 = 10.
You're right to say that it isn't popular. I guess I only saw it in
one small place where they were explaining inductive reasoning and
created a notation to save themselves from saying "1+2+...+n" any more
than necessary.
I suppose it would be more common if you couldn't say the same thing
as C(n,2) -- it's just as intuitive as factorial, and probably comes
up as often.
>Now one operator that I have seen (Gradshteyn uses it) is the bifactorial
>operator, '!!', where n!! = product n(n-2)(n-4)... down to 2 or 1.
>Thus 4!! = 8 and (Gamma(4))!! = 48.
The increment operator is also very popular, doesn't mean that it should
be added to the set. :-)
This really puts me in the mind to write a program that would accept
a variable set of operations and calculates which numbers it can form
with so many 4's. For instance, if you included exponentiation, then
for all X and Y that you could form with #X and #Y 4's, you could
form X^Y with #(X^Y) = #X + #Y. If you had a bifactorial operation,
then for any X you could create X!! and #(X!!) = #X. And if you had
repeating decimals, then you allow 4/9 as a primitive that you can
form with #(4/9) = 1.
I suspect that such a program would have to be clever about creating
bounds for what numbers it would catalog. I mean, you CAN form an
infinite number of numbers with just one 4, although only a very finite
number of them are relevant to forming numbers from 1-100. Hmmmm.
Abigail
Wei-Hwa Huang (whu...@ugcs.caltech.edu) wrote on 1386 September 1993 in
<URL: news:5o67qd$a...@gap.cco.caltech.edu>:
++
++ ObPuzzle: What was the first publication to ever accept an article
++ written by Donald Knuth? The answer may surprise you.
MAD Magazine.
Kunth, D. E.: "The potrzebie system of weights and measures."
MAD Magazine 33 (June 1957), 36--37. (Illustrated by Wallace Wood.)
Reprinted in: "Like, MAD" (New York: Signet Pocket Books No. S1838, 1960),
139--145. Page 36 reprinted in: "Completely MAD", by Maria Reidelbach
(Boston, Mass.: Little, Brown, 1991), 191.
[http://www-cs-faculty.Stanford.EDU/%7Eknuth/vita.tex.gz]
Abigail (yeah, I read the TeX source, got a problem with that?)
Fred the Wonder Worm
In article <5o3lhl$bk6$1...@kodak.rdcs.kodak.com>,
Matthew Daly <da...@PPD.Kodak.COM> wrote:
>
>ObPuzzle: Give yourself gamma and triangle, but eliminate the "base-10"
>operations. Can you form 9, 10, and 44 with one or two 4's? Would that
>make the "baseless" system equally powerful?
As someone has already observed, 10 = T(4) and 9 = T(4) - G(S(4))
(where T, G and S stand for Triangle, Gamma and Sqrt), but I can't
do 44 with only two fours.
ObPuzzle: With the same restrictions, find an expression for 35 using
only two 4's.
Cheers,
Geoff.
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Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
ft...@cs.su.oz.au | Gameplayer by vocation.
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