Four 4s ongoing challange!

RyanJ · October 15, 2005

Seeing as no-one has posted for a few days I'd say the challange has finished which means the winner was cosine who had 119!

Congradulations to you all - this is the highest result I have seen on any forum to date

Cheers,

Ryan Jones

NPK · December 25, 2005

(√4 / .4)! + 4 - 4 = 120

(44/4)^ √4 = 121

(√4 / .4)! + 4 - √4 = 122

cosine · December 25, 2005

(√4 / .4)! + 4 - 4 = 120

(44/4)^ √4 = 121

(√4 / .4)! + 4 - √4 = 122

Wowzers! I was just looking at this thread last night too! (My first SFN thread, sigh...)

RyanJ · December 25, 2005

Hmm, looks like the challange has been re-opened then - fair enough

Cheers,

Ryan Jones

cosine · December 25, 2005

Hmm' date=' looks like the challange has been re-opened then - fair enough

Cheers,

Ryan Jones[/quote']

Yeah... so did we ever decide what functions should be allowed?

RyanJ · December 25, 2005

Yeah... so did we ever decide what functions should be allowed?

I belive I did say any function in the original post:

3. You may use any mathematical operations and symbols (not including symbols for other numbers' date=' like Pi, e, etc.) you wish.

[/quote']

I think the bold would allow for functions as long as your uing only 4's as the values I see no probles there

Cheers,

Ryan Jones

NPK · December 26, 2005

(√√√(√4/.4)^4!) - √4 = 123

or

44/.4' + 4! = 123 (.4' = .4 recurring)

Trond · December 26, 2005

I saw someone posting this when I made a similar thread on a math-forum:

4/4=1

4+4/4=2

4+4+4/4=3

4+4+4+4/4=4

4+4+4+4+4/4=5

4+4+4+4+4+4/4=6

4+4+4+4+4+4+4/4=7

4+4+4+4+4+4+4+4/4=8

4+4+4+4+4+4+4+4+4/4=9

4+4+4+4+4+4+4+4+4+4/4=10

There are as many fours adding together as the number is in the answer,

and it seems to go on forever

the tree · December 26, 2005

Isn't that a little obvious?

[math]\frac{ab}{b}=a[/math]

BigMoosie · December 27, 2005

A while back I continued with my own list using only basic algebraic expressions (+ - / * ^ ! etc), and I managed to get all but three numbers less than 100. Anyone wanna help get 89, 91, 93 using these rules?

An underline indicates repetition: (i.e: .4 = .44444 = 4/9):

1 = 4*4/(4*4)
2 = 4/4+4/4
3 = (4+4+4)/4
4 = (4-4)/4+4
5 = 4^(4-4)+4
6 = (4+4)/4+4
7 = 4+4-4/4
8 = 4+4+4-4
9 = 4/4+4+4
10 = (4*4+4!)/4
11 = (4+4!)/4+4
12 = (4-4/4)*4
13 = (4+4!+4!)/4
14 = 4!/4+4+4
15 = 4*4-4/4
16 = 4*4+4-4
17 = 4*4+4/4
18 = (4*4!-4!)/4
19 = 4!-(4+4/4)
20 = (4/4+4)*4
21 = 4!+4/4-4
22 = 4!-(4+4)/4
23 = 4!-4^(4-4)
24 = 4*4+4+4
25 = 4!+(4/4)^4
26 = 4!+4!/4-4
27 = 4!+4-4/4
28 = (4+4)*4-4
29 = 4/4+4!+4
30 = (4*4!+4!)/4
31 = (4+4!)/4+4!
32 = 4^4/(4+4)
33 = (4-.4)/.4+4!
34 = 4!/4+4+4!
35 = (4.4/.4)+4!
36 = (4+4)*4+4
37 = 4/.[u]4[/u]+4+4!
38 = 44-4!/4
39 = (4*4-.4)/.4
40 = (4^4/4)-4!
41 = (4*4+.4)/.4
42 = 4!+4!-4!/4
43 = 44-4/4
44 = 4*4+4+4!
45 = (4!/4)!/(4*4)
46 = (4!-4)/.4 - 4
47 = 4!+4!-4/4
48 = (4*4-4)*4
49 = 4!+4!+4/4
50 = (4*4+4)/.4
51 = 4!/.4-4/.[u]4[/u]
52 = 44+4+4
53 = 44+4/.[u]4[/u]
54 = (4!/4)^4/4!
55 = (4!-.4)/.4-4
56 = 4!+4!+4+4
57 = 4/.[u]4[/u]+4!+4!
58 = (4^4-4!)/4
59 = 4!/.4-4/4
60 = 4*4*4-4
61 = 4!/.4+4/4
62 = (4!+.4+.4)/.4
63 = (4^4-4)/4
64 = 4^(4-4/4)
65 = 4^4+4/4
66 = (4+4!)/.4-4
67 = (4+4!)/.[u]4[/u]+4
68 = 4*4*4+4
69 = (4+4!-.4)/.4
70 = (4^4+4!)/4
71 = (4!+4.4)/.4
72 = (4-4/4)*4!
73 = ([sup].4[/sup]√4+.[u]4[/u])/.[u]4[/u]
74 = (4+4!)/.4+4
75 = (4!/4+4!)/.4
76 = (4!-4)*4-4
77 = (4!-.[u]4[/u])/.[u]4[/u]+4!
78 = (4!x.[u]4[/u]+4!)/.[u]4[/u]
79 = ([sup].4[/sup]√4-.4)/.4
80 = (4*4+4)*4
81 = (4/4-4)^4
82 = 4!/.[u]4[/u]+4!+4
83 = (4!-.4)/.4+4!
84 = (4!-4)*4+4
85 = (4/.4+4!)/.4
86 = (4-.4)x4!-.4
87 = 4!x4-4/.[u]4[/u]
88 = 4^4/4+4!
[size="4"][color="Red"][b]89[/b][/color][/size]
90 = (4!/4)!/(4+4)
[size="4"][color="Red"][b]91[/b][/color][/size]
92 = (4!-4/4)*4
[size="4"][color="Red"][b]93[/b][/color][/size]
94 = (4+4!)/.4 + 4!
95 = 4!*4-4/4
96 = 4!*4+4-4
97 = 4!*4+4/4
98 = (4!+.4)*4+.4
99 = (4!+4!-4)/.[u]4[/u]
100 = 4*4/(.4*.4)

Here is a reason why in my opinion using any functions is pointless:

[math]0 = \log_{\tfrac{4}{.4}}\frac{4}{4}[/math]

[math]1 = \log_{\tfrac{4}{.4}}\frac{4}{.4}[/math]

[math]2 = \log_{\tfrac{4}{.4}}\frac{4}{4\%}[/math]

[math]3 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%}[/math]

[math]4 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%}[/math]

[math]5 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%}[/math]

[math]6 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%}[/math]

[math]7 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%\%}[/math]

[math]8 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%\%}[/math]

[math]9 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%\%\%}[/math]

[math]10 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%\%\%}[/math]

cosine · December 27, 2005

A while back I continued with my own list using only basic algebraic expressions (+ - / * ^ ! etc)' date=' and I managed to get all but three numbers less than 100. Anyone wanna help get 89, 91, 93 using these rules?

An underline indicates repetition: (i.e: .[u']4[/u] = .44444 = 4/9):
1 = 4*4/(4*4)
2 = 4/4+4/4
3 = (4+4+4)/4
4 = (4-4)/4+4
5 = 4^(4-4)+4
6 = (4+4)/4+4
7 = 4+4-4/4
8 = 4+4+4-4
9 = 4/4+4+4
10 = (4*4+4!)/4
11 = (4+4!)/4+4
12 = (4-4/4)*4
13 = (4+4!+4!)/4
14 = 4!/4+4+4
15 = 4*4-4/4
16 = 4*4+4-4
17 = 4*4+4/4
18 = (4*4!-4!)/4
19 = 4!-(4+4/4)
20 = (4/4+4)*4
21 = 4!+4/4-4
22 = 4!-(4+4)/4
23 = 4!-4^(4-4)
24 = 4*4+4+4
25 = 4!+(4/4)^4
26 = 4!+4!/4-4
27 = 4!+4-4/4
28 = (4+4)*4-4
29 = 4/4+4!+4
30 = (4*4!+4!)/4
31 = (4+4!)/4+4!
32 = 4^4/(4+4)
33 = (4-.4)/.4+4!
34 = 4!/4+4+4!
35 = (4.4/.4)+4!
36 = (4+4)*4+4
37 = 4/.[u]4[/u]+4+4!
38 = 44-4!/4
39 = (4*4-.4)/.4
40 = (4^4/4)-4!
41 = (4*4+.4)/.4
42 = 4!+4!-4!/4
43 = 44-4/4
44 = 4*4+4+4!
45 = (4!/4)!/(4*4)
46 = (4!-4)/.4 - 4
47 = 4!+4!-4/4
48 = (4*4-4)*4
49 = 4!+4!+4/4
50 = (4*4+4)/.4
51 = 4!/.4-4/.[u]4[/u]
52 = 44+4+4
53 = 44+4/.[u]4[/u]
54 = (4!/4)^4/4!
55 = (4!-.4)/.4-4
56 = 4!+4!+4+4
57 = 4/.[u]4[/u]+4!+4!
58 = (4^4-4!)/4
59 = 4!/.4-4/4
60 = 4*4*4-4
61 = 4!/.4+4/4
62 = (4!+.4+.4)/.4
63 = (4^4-4)/4
64 = 4^(4-4/4)
65 = 4^4+4/4
66 = (4+4!)/.4-4
67 = (4+4!)/.[u]4[/u]+4
68 = 4*4*4+4
69 = (4+4!-.4)/.4
70 = (4^4+4!)/4
71 = (4!+4.4)/.4
72 = (4-4/4)*4!
73 = ([sup].4[/sup]√4+.[u]4[/u])/.[u]4[/u]
74 = (4+4!)/.4+4
75 = (4!/4+4!)/.4
76 = (4!-4)*4-4
77 = (4!-.[u]4[/u])/.[u]4[/u]+4!
78 = (4!x.[u]4[/u]+4!)/.[u]4[/u]
79 = ([sup].4[/sup]√4-.4)/.4
80 = (4*4+4)*4
81 = (4/4-4)^4
82 = 4!/.[u]4[/u]+4!+4
83 = (4!-.4)/.4+4!
84 = (4!-4)*4+4
85 = (4/.4+4!)/.4
86 = (4-.4)x4!-.4
87 = 4!x4-4/.[u]4[/u]
88 = 4^4/4+4!
[size="4"][color="Red"][b]89[/b][/color][/size]
90 = (4!/4)!/(4+4)
[size="4"][color="Red"][b]91[/b][/color][/size]
92 = (4!-4/4)*4
[size="4"][color="Red"][b]93[/b][/color][/size]
94 = (4+4!)/.4 + 4!
95 = 4!*4-4/4
96 = 4!*4+4-4
97 = 4!*4+4/4
98 = (4!+.4)*4+.4
99 = (4!+4!-4)/.[u]4[/u]
100 = 4*4/(.4*.4)
Here is a reason why in my opinion using any functions is pointless:

[math]0 = \log_{\tfrac{4}{.4}}\frac{4}{4}[/math]

[math]1 = \log_{\tfrac{4}{.4}}\frac{4}{.4}[/math]

[math]2 = \log_{\tfrac{4}{.4}}\frac{4}{4\%}[/math]

[math]3 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%}[/math]

[math]4 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%}[/math]

[math]5 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%}[/math]

[math]6 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%}[/math]

[math]7 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%\%}[/math]

[math]8 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%\%}[/math]

[math]9 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%\%\%}[/math]

[math]10 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%\%\%}[/math]

Good arguement, I think.

RyanJ · December 27, 2005

Good arguement, I think.

If you can think of a rule that allows functions and stops them being "abused" then post it

Cheers,

Ryan Jones

cosine · December 27, 2005

If you can think of a rule that allows functions and stops them being "abused" then post it

Cheers' date='

Ryan Jones[/quote']

Thats the problem, its hard to differentiate one function from another (pun intended).

Although, I was thinking, how about this:

If a function can be defined by a differential equation, then it should be accepted. And if a function can only be defined by a difference equation, then it shouldn't be accepted.

Though it seems a nice enough rule, it allows for what Big moosie's arguement, so...?

Edit: I don't like this rule already, cause it would allow:

[math]f(x) = N[/math]

thus,

[math]f(4+4+4+4) = N[/math]

cosine · December 29, 2005

Thats the problem' date=' its hard to differentiate one function from another (pun intended).

Although, I was thinking, how about this:

If a function can be defined by a differential equation, then it should be accepted. And if a function can only be defined by a difference equation, then it shouldn't be accepted.

Though it seems a nice enough rule, it allows for what Big moosie's arguement, so...?

Edit: I don't like this rule already, cause it would allow:

[math']f(x) = N[/math]

thus,

[math]f(4+4+4+4) = N[/math]

There's always the 4 or 5 operator rule, where only + - * / and ^ (and parenthesis) are allowed...

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