Four Fours = 33?
Peter Kirby
I have been puzzling over this for a while. Please help!
There is a certain math teacher of mine who holds that all positive
integers can be written as an expression with only mathematical symbols
and four 4's. For example,
1: (4 + 4) / (4 + 4)
2: 4 / 4 + 4 / 4
3: (4 + 4 + 4) / 4
Et cetera. Just to make sure your following along, find a Four Fours
for 31.
(Hint: You can use +, -, *, /, !, and sqrt. 4! = 4*3*2*1 = 24.)
Before we get to my questions, we must first consider which
operators/functions we can use. I suggest that we can use anything but
a rounding function. That means +, -, *, /, sqrt, !, ^, (, ), ., log,
sin, cos, tan, and anything else you can think of (that won't let you
build a rounding function). You can use some fours to make a decimal
number (i.e., use two fours to make 44, 4.4, or .44). Of course, you
can only use the digit four, it can only appear four times, and you must
use base decimal.
It isn't difficult up to 32, but 33 is a stumbling block. Can you
A) Find a Four Fours expression for 33?
B) Explain why you can't?
In general, can you demonstrate that
A1) Every whole number has a Four Fours expression?
A2) Given a rounding function, every whole number has a Four Fours
expression?
B1) There is a whole number that doesn't have a Four Fours expression?
B2) Even with a rounding function, there isn't a Four Fours expression
for every whole number?
Any insights appreciated,
Peter Kirby
PS-
Perhaps this will save you a little grief. To try to answer my
questions, I wrote a program to generate a subset of Four Fours
expressions. It found the combinations of the operators +, -, *, /, and
^ on four numbers 2, 4, and 24. (2 is sqrt(4) and 24 is 4!)
That is, it would find:
(4 + (24 / 24)) ^ 2 = 25
but it wouldn't find
(sqrt(24) ^ 2) + (4 - 4) = 24
because sqrt isn't one of the operators used by the program.
Only a finite number of expressions can be found this way, but that was
what I wanted. After all these expressions were found, the program
printed out an expression (if any) that it found for each number 1 to
1000. The first number that was not discovered was 33, and most of the
larger numbers were also not found. This doesn't really show anything
except that anyone who wants to find 33 with Four Fours will have to be
pretty creative.
Tim Firman
Tim Firman wrote:
> I didn't find one, but didn't prove it's impossible. A different
> version of this puzzle, using 'buttons on a scientific calculator'
> would make this easy; 4[x^2]+ 4[x^2] +4/4
I fiddled with this again, and found an answer requiring only four
fours, but an infinite number of uses for a function.
4! + (4/.4) -(sqrt(sqrt(sqrt(.......ad infinitum(4)
for 24 + 10 - 1 = 33
Tim Firman
Peter Kirby wrote:
> There is a certain math teacher of mine who holds that all positive
> integers can be written as an expression with only mathematical symbols
> and four 4's. For example,
> 1: (4 + 4) / (4 + 4)
> 2: 4 / 4 + 4 / 4
> 3: (4 + 4 + 4) / 4
> Before we get to my questions, we must first consider which
> operators/functions we can use. I suggest that we can use anything but
> a rounding function. That means +, -, *, /, sqrt, !, ^, (, ), ., log,
> sin, cos, tan, and anything else you can think of (that won't let you
> build a rounding function). You can use some fours to make a decimal
> number (i.e., use two fours to make 44, 4.4, or .44). Of course, you
> can only use the digit four, it can only appear four times, and you must
> use base decimal.
> It isn't difficult up to 32, but 33 is a stumbling block. Can you
> A) Find a Four Fours expression for 33?
> B) Explain why you can't?
I didn't find one, but didn't prove it's impossible. A different
version of this puzzle, using 'buttons on a scientific calculator'
would make this easy; 4[x^2]+ 4[x^2] +4/4
> In general, can you demonstrate that
> A1) Every whole number has a Four Fours expression?
> A2) Given a rounding function, every whole number has a Four Fours
> expression?
> B1) There is a whole number that doesn't have a Four Fours expression?
> B2) Even with a rounding function, there isn't a Four Fours expression
> for every whole number?
All of these have a simple answer. Given a rigidly defined set of
rules of how to use functions, there will be a finite number of
combinations of the four 4's and a finite number of allowed functions.
This can specify only a finite number of numbers, while there are an
infinite number of whole numbers.
This requires specifying a finite number of functions; if I get to use
an infinite number of functions, well, I can think of the function
F(x) = 33
giving the answer:
F(4) + 4 - Sqrt(4) - Sqrt(4) = 33
And so on, using a different function for each number.
I don't think that was what was intended.
Tim
hyal...@mit.edu
> All of these have a simple answer. Given a rigidly defined set of
>rules of how to use functions, there will be a finite number of
>combinations of the four 4's and a finite number of allowed functions.
>This can specify only a finite number of numbers, while there are an
>infinite number of whole numbers.
Not necessarily. If any of the functions is unary, then you could get an
infinite number of possible values. For example, 4!, (4!)!, ((4!)!)!,...
Hareendra Yalamanchili
hyal...@mit.edu
William and Alice Englander
In article <326187...@earthlink.net>, Peter Kirby <ki...@earthlink.net> writes:
>
> It isn't difficult up to 32, but 33 is a stumbling block. Can you
> A) Find a Four Fours expression for 33?
> B) Explain why you can't?
>
If I understand the question correctly, the following will
create the value 33 using 4 4's:
[SPOILER]
4 - .4
4! + -------- = 33
.4
Bill E.
--
William and Alice Englander
engl...@netcom.com
http://www.englander.com
Rolf Strebel
> There is a certain math teacher of mine who holds that all positive
> integers can be written as an expression with only mathematical symbols
> and four 4's. For example,
> 1: (4 + 4) / (4 + 4)
> 2: 4 / 4 + 4 / 4
> 3: (4 + 4 + 4) / 4
> Before we get to my questions, we must first consider which
> operators/functions we can use. I suggest that we can use anything but
> a rounding function. That means +, -, *, /, sqrt, !, ^, (, ), ., log,
> sin, cos, tan, [...]
> It isn't difficult up to 32, but 33 is a stumbling block. Can you
> A) Find a Four Fours expression for 33?
If you're allowed to use the log base 2 (logb), it's easy
1 = logb(logb(4))
2 = logb(4)
33 = 4*(4 + 2) + 1
Furthermore, if ! may be applied to all real numbers, then
expressions like
log(log(4!)!)
give you lots of real numbers.
-- Rolf
Sasha Semenov
> integers can be written as an expression with only mathematical symbols
> and four 4's. For example,
> 1: (4 + 4) / (4 + 4)
> 2: 4 / 4 + 4 / 4
> 3: (4 + 4 + 4) / 4
>
> for 31.
> (Hint: You can use +, -, *, /, !, and sqrt. 4! = 4*3*2*1 = 24.)
>
> operators/functions we can use. I suggest that we can use anything but
> a rounding function. That means +, -, *, /, sqrt, !, ^, (, ), ., log,
> build a rounding function). You can use some fours to make a decimal
> number (i.e., use two fours to make 44, 4.4, or .44). Of course, you
> can only use the digit four, it can only appear four times, and you must
> use base decimal.
SPOILER
Any integer may be expessed using 3 fours:
log log sqrt(sqrt(sqrt(...sqrt(4)...))) =
4 4 n times
= log log 4^(1/2^n) = log 1/2^n = -n/2
4 4 4
Susan Hoover
Tim Firman <fir...@lombardi.chem.wisc.edu> wrote:
> I fiddled with this again, and found an answer requiring only four
>fours, but an infinite number of uses for a function.
>
> 4! + (4/.4) -(sqrt(sqrt(sqrt(.......ad infinitum(4)
>
>for 24 + 10 - 1 = 33
Looking at your method, and thinking a bit, I get
4! + (4 - .4)/.4 = 24 + 3.6/.4 = 24 + 9 = 33
Peter Kirby
Thank you all for your kind help! My questions have been answered, so
let this thread die. Finding 33 is easier than I thought. As Bill E.
said,
4 - .4
33 = --------- + 4! = 9 + 24 = 33
.4
And Sasha Semenov was on the right track when she said that all whole
numbers can be found using log's and a series of sqrt's. I got the
following letter:
> Let X = sqrt(sqrt(sqrt(sqrt(....(sqrt(4)))) (29 sqrts in total)
>
> Then log( log(4)/log(X) )/log(sqrt(4)) + 4 is 33.
>
> This trick works for any whole number>4; just use n-4 sqrts.
>
> Note that log(4)/log(X) = 2^n, where n is the number of sqrts.
> Then log(2^n)/log(2) = n.
> We still have a 4 left over, so I just add it on and reduce the number of
> sqrts.
>
> Ken Shirriff shir...@eng.sun.com http://www.cs.berkeley.edu/~shirriff
rec.puzzles: Your 24-hour information delivery service. =)
Sincerely,
Peter Kirby
Tim Firman
Susan Hoover wrote:
> 4! + (4 - .4)/.4 = 24 + 3.6/.4 = 24 + 9 = 33
Thanks. I'm not sure why I failed to see that...
Fred the Wonder Worm
In article <STREBEL.96...@ru7.inf.ethz.ch>,
Rolf Strebel <str...@ru7.inf.ethz.ch> wrote:
>> There is a certain math teacher of mine who holds that all positive
>> integers can be written as an expression with only mathematical symbols
>> and four 4's. For example,
>> 1: (4 + 4) / (4 + 4)
>> 2: 4 / 4 + 4 / 4
>> 3: (4 + 4 + 4) / 4
>>
>> Before we get to my questions, we must first consider which
>> operators/functions we can use. I suggest that we can use anything but
>> a rounding function. That means +, -, *, /, sqrt, !, ^, (, ), ., log,
>> sin, cos, tan, [...]
>>
>> It isn't difficult up to 32, but 33 is a stumbling block. Can you
>> A) Find a Four Fours expression for 33?
>
>If you're allowed to use the log base 2 (logb), it's easy
>
>1 = logb(logb(4))
>2 = logb(4)
>33 = 4*(4 + 2) + 1
Given log, sqrt, - and / it is possible for all integers n. If the log
given requires the base specified, then the following will do:
n = -log_4 (log_4 (sqrt (sqrt ( ... (sqrt (4*4)) ... ))))
where there are 2*n+1 sqrts (this is for non-negative n; negative n is left
as an exercise for the reader :).
If the log function supplied has a specific base (e or 10 being most likely),
then the following will do:
n = [ log (log 4) - log (log (sqrt ... (sqrt (4*4)) ... )) ] / log(4)
where there are 2*n+1 sqrts for non-negative n. Once again negative n is
left as a trivial exercise. For this reason, log is usually disallowed in
such problems, although the problem is really the sqrt function. But sqrt
is seen as a more 'natural' operation to allow.
Cheers,
Geoff.
-------------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
ft...@cs.su.oz.au | Gameplayer by vocation.
-------------------------------------------------------------------------------
Patrick Hamlyn
Tim Firman <fir...@lombardi.chem.wisc.edu> wrote:
>Tim Firman wrote:
>> I didn't find one, but didn't prove it's impossible. A different
>> version of this puzzle, using 'buttons on a scientific calculator'
>> would make this easy; 4[x^2]+ 4[x^2] +4/4
> I fiddled with this again, and found an answer requiring only four
>fours, but an infinite number of uses for a function.
> 4! + (4/.4) -(sqrt(sqrt(sqrt(.......ad infinitum(4)
>for 24 + 10 - 1 = 33
This is easily fixed without the infinite functions stuff:
.
4! + (4/.44) = 33
That's .44 recurring, ie 4/9
I got all these once up to about 87.
+=======================================+
| Name: Patrick Hamlyn |
| Company: Multiprogramming Pty Ltd |
| Email: pa...@multipro.com.au |
+=======================================+
Andrew Nikitin
14 Oct 96, Peter Kirby wrote:
PK> I suggest that we can use anything
PK> but a rounding function. That means +, -, *, /, sqrt, !, ^, (, ), .,
^ ^^^^ ^^^^
PK> log, sin, cos, tan, and anything else you can think of (that won't
^^^
PK> let you build a rounding function).
PK> It isn't difficult up to 32, but 33 is a stumbling block. Can you
????????? Really? Do you know that using operations above you can express ANY
whole number using only three two's? Literally
n=-log log √√...√2
2 2 ^^^^^^ n times
(√ symbol looks like "sqrt(" on my display)
PK> A) Find a Four Fours expression for 33?
With very little changes this method may be applied to 4 fours:
n=-log log √√...√4
4 4 ^^^^^^ 2*n times
You can say that only 3 fours were used in this expression. Ok, replace one
of the fours by √4^√4 or something like this. Also use 66 sqrt signs to get 33.
PK> B) Explain why you can't?
Why "can't"? I can.
PK> A2) Given a rounding function, every whole number has a Four Fours
PK> expression?
puzzle: Given a rounding function every whole number has "one x" expression
(where x is any). Prove it.
--- NSG
* Origin: (2:461/44.26)
Arthur C Clay
|> Peter Kirby wrote:
|> > There is a certain math teacher of mine who holds that all positive
|> > integers can be written as an expression with only mathematical symbols
|> > and four 4's. For example,
|> > 1: (4 + 4) / (4 + 4)
|> > 2: 4 / 4 + 4 / 4
|> > 3: (4 + 4 + 4) / 4
|>
|> > Before we get to my questions, we must first consider which
|> > operators/functions we can use. I suggest that we can use anything but
|> > a rounding function. That means +, -, *, /, sqrt, !, ^, (, ), ., log,
|> > sin, cos, tan, and anything else you can think of (that won't let you
|> > build a rounding function). You can use some fours to make a decimal
|> > number (i.e., use two fours to make 44, 4.4, or .44). Of course, you
|> > can only use the digit four, it can only appear four times, and you must
|> > use base decimal.
|>
|> > It isn't difficult up to 32, but 33 is a stumbling block. Can you
|> > A) Find a Four Fours expression for 33?
|> > B) Explain why you can't?
A) arctan(4/4) - 4!/sqrt(4) = 45 - 24/2 = 45 - 12 = 33
B) ?
Sharon Brindle
The puzzle as I was told it did not allow any "word" functions
such as log sin cos etc. Only the symbolic functions such as the
exponents, multiplication, square root etc are allowed. This
makes it a whole lot more difficult.
Now if anyone can make four fours equal 73 given this limitation
I would be extremely grateful, having managed to get every number
from 1 to 100 bar this one...
--
Sharon Brindle
Aaron Liang
> I have been puzzling over this for a while. Please help!
> integers can be written as an expression with only mathematical symbols
> and four 4's. For example,
> 1: (4 + 4) / (4 + 4)
> 2: 4 / 4 + 4 / 4
> 3: (4 + 4 + 4) / 4
> for 31.
> (Hint: You can use +, -, *, /, !, and sqrt. 4! = 4*3*2*1 = 24.)
> operators/functions we can use. I suggest that we can use anything but
> a rounding function. That means +, -, *, /, sqrt, !, ^, (, ), ., log,
> sin, cos, tan, and anything else you can think of (that won't let you
> build a rounding function). You can use some fours to make a decimal
> number (i.e., use two fours to make 44, 4.4, or .44). Of course, you
> can only use the digit four, it can only appear four times, and you must
> use base decimal.
> It isn't difficult up to 32, but 33 is a stumbling block. Can you
> A) Find a Four Fours expression for 33?
> B) Explain why you can't?
> B1) There is a whole number that doesn't have a Four Fours expression?
> Any insights appreciated,
> Peter Kirby
> PS-
> Perhaps this will save you a little grief. To try to answer my
> questions, I wrote a program to generate a subset of Four Fours
> expressions. It found the combinations of the operators +, -, *, /, and
> ^ on four numbers 2, 4, and 24. (2 is sqrt(4) and 24 is 4!)
> That is, it would find:
> (4 + (24 / 24)) ^ 2 = 25
> but it wouldn't find
> (sqrt(24) ^ 2) + (4 - 4) = 24
> because sqrt isn't one of the operators used by the program.
> Only a finite number of expressions can be found this way, but that was
> what I wanted. After all these expressions were found, the program
> printed out an expression (if any) that it found for each number 1 to
> 1000. The first number that was not discovered was 33, and most of the
> larger numbers were also not found. This doesn't really show anything
> except that anyone who wants to find 33 with Four Fours will have to be
> pretty creative.
Mr Bruce
= 30 + 24 / 8 = 30 + 3