The game of four 4s requires you to construct a given number by
combining exactly four 4s using any conventional mathematical symbols.

Rules:

1.  Any conventional symbols are allowed (and not only + - x and / as
in the simple version of the game)

2.  Functions requiring the use of letters such as sin log etc are not
allowed.

3.  No constants other than 4 may be used (eg no e or pi)

Examples:

71  =  (4! + 4.4)/.4
ie (4 factorial plus 4 point 4)divided by point 4

          __           .
111 =  (\/4 /.4)! - 4/.4
ie factorial(root 4 over point 4) minus (4 over point 4 recurring)

                 .
131  = 4!/(.4 x .4) -4
ie 4 factorial over (point 4 times point 4 recurring) minus 4

The question:

Using the above rules, make exactly 107.

.      .
(4! + 4! - .4) / .4

I found this solution at http://www.wheels.org/math/44s.html

Search term used on Google:
"four fours problem" 107

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Clarification of Answer by blazius-ga on 23 Aug 2002 03:30 PDT

The formatting of my answer got garbled, the points in the first line
should be over the last .4's in the second line.  This represents the
follwing calculation):

(4! + 4! - .444444444444444444...) / .444444444444444444... = 107

---

Request for Answer Clarification by michael2-ga on 23 Aug 2002 04:26 PDT

In the web site you found, the author states:

"It can be proved that not all integers can be represented."

How?

(ps no prizes for that one; just out of interest)

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Clarification of Answer by blazius-ga on 23 Aug 2002 04:42 PDT

I really don't know.  I am not a mathematician, nor have I been able
to find any easily accessable proof for this claim.  I suggest that
you post this as a seperate question.

Very good!  I'd disqualify some of the answers given by your source,
though, for using non-permitted zeros, eg the .04 which is used in the
solution to 103.  Better would be:

           .
103  = 44/.4 + 4

I can prove that not all integers can be represented.

It's this simple. Find the LARGEST number you can make with the 4's
(I'm not far enough into life to know tha tmuch math, but something
like
4!^4!^(4!^4!)

Now you cannot make a number bigger can you?

Could you make 99x10^1,000,000?
I doubt it - the numbers just don't get big enough.

I hope that helps :)
-Matt

To murph-ga:

Well, you could add !'s ad infinitum:

4!!!!!!!!!!!!!!!!!!!!!!!!!^4!!!!!!!!!!!!!!!!!!!!!!!!!!!!!^(4!!!!!!!!!!!!!!!!!!!!!!!!!!!!!^4!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!)

and get terribly large numbers.

if it's called the game of four 4s the answer given for 107 shouldn't be allowed

blazius-ga

Or even better, you can do the following:

4/(4-4)

I tried solving without looking at the answer and I didn't think of
the dot notation for a decimal of repeating 4's.  On the other hand, I
did use the floor (first integer less than a real) and ceil (first
integer greater than a real) functions.  I consider those to be at
least as common as the repeating dot notation and there are well
defined mathematical symbols for them.

Using floor and ceil, there are lots of values that can be easily made
- i.e. floor(sqrt(sqrt(4))) = 1, ceil(sqrt(4!)) = 5, floor(sqrt(44)) =
6, ceil(sqrt(44)) = 7.

Given those, I was pretty sure that I could get to 107 by taking the
floor of some number of square roots of some number or factorial of 4
or any of the other numbers I can make (i.e.
floor(sqrt(sqrt(...(sqrt((4!)!))...)) I haven't tried to figure this
out but it seems promising to me.  I would also suspect that with ceil
and floor it may be possible to make all possible integers.

Naturally, all this is a slippery slope since mathematician are always
creating new operations. There is nothing to prevent them (or even me)
from creating a new operation that takes four 4s and produces 107.

Steve

p.s. I did reject the ++ operator which increments its argument by 1
(i.e. 4++ = 5), since that would make the problem too easy.