Classic problem: Four 4's
 

I found a version of the four 4's problem in an old book awhile back. [i]Four[/i] those who haven't seen it be[i]four[/i], the challenge is to create [i]four[/i]mulas, using no more than [i]four[/i] 4's, that evaluate to other integers. :grin:

This version is more stringent on the operations allowed than most. The operations allowed are listed below. The book claims that all positive integers up to 119 can be created this way. I have been unable to create three of those integers, so I'd like to see what you can come up with. If the book is correct, though, all integers from 1 through at least 130 can be created.

Allowed unary operations: (each example in parentheses uses one 4)
Square (4^2) - this is the only place where 2's are allowed. Also note below that exponentiation other than this is [b]not[/b] allowed.
Factorial (4!)
Decimal (.4)

Allowed binary operations: (each example in parentheses uses two 4's)
+ (4+4)
- (4-4)
* (4*4)
/ (4/4)
Concatenate (44); also before (4.4) or after (.44) a Decimal.
Any number of parentheses are also allowed. Standard order of operations applies when parentheses are missing.

Since it's usually easy to fill out an expression with more 4's (e.g. 8 = 4+4 or 4^2/4+4^2/4), I'll determine best solutions first by fewest 4's, and second by shortest total length.

Wacky, would you mind setting up a scoreboard from 0 to 160 or so? I'll start populating it:

[code]N: 4's,len expression (Author, date)
8: 2, 3: 4+4 (Ken_g6, 9/28)[/code]

But I won't do any more unless you guys get desperate. :wink:

[QUOTE=Ken_g6]This version is more stringent on the operations allowed than most.[/quote]I've never seen another version that allowed squaring as a unary operator. OTOH versions that I've seen allowed floor, ceiling, square root and exponentiation, so this'll be more difficult in some cases.

Notation: I'll use 4[sup]sq[/sup] to denote the unary squaring, so as not to introduce any "2".

Here's some easy or obvious ones, not necessarily of minimal length, just to initiate a few table entries:

4 = 4

6 = 4! / 4

10 = 4 / .4
11 = 44 / 4

16 = 4[sup]sq[/sup]

24 = 4!

32 = 4[sup]sq[/sup] + 4[sup]sq[/sup]

36 = 4![sup]sq[/sup] / 4[sup]sq[/sup]

44 = 44

48 = 4! + 4!

64 = 4[sup]sq[/sup] * 4

72 = 4! * 4 - 4!

80 = 4[sup]sq[/sup] * 4 + 4[sup]sq[/sup]

96 = 4! * 4

100 = ( 4 / .4 )[sup]sq[/sup]

110 = 44 / .4
111 = 44.4 / .4

128 = 4[sup]sq[/sup] * 4 + 4[sup]sq[/sup] * 4

144 = 4![sup]sq[/sup] / 4

256 = 4[sup]sq[/sup] * 4[sup]sq[/sup]

[QUOTE=cheesehead]256 = 4[sup]sq[/sup] * 4[sup]sq[/sup][/QUOTE]256 = 4[sup]sq[/sup][sup]sq[/sup]

I'll set up a scoreboard soon.

Obviously:
0 = 4-4
1 = 4/4
2 = (4+4)/4
3= (4+4+4)/4
5 = 4 + 4/4
7 = 4 + 4 - 4/4
8 = 4 + 4
9 = 4 + 4 + 4/4
12 = 4 + 4 + 4

13 = 4sq -4 + 4/4
15 = 4sq - 4/4

[quote=ken_g6]The book claims that all positive integers up to [B]119[/B] can be created this way[/quote]
[quote=ken_g6]If the book is correct, though, all integers from 1 through at least [B]130[/B] can be created.[/quote]
[quote=ken_g6] Wacky, would you mind setting up a scoreboard from 0 to [B]160[/B] or so?[/quote]
[quote=cheesehead][B]256[/B] = 4[sup]sqsq[/sup][/quote]

Did I miss something ?

17 = 4sq + 4/4
18 = 4/.4 + 4 + 4
19 = 4! - 4 - 4/4
20 = 4! - 4
21 = 4! -4 + 4/4
23 = 4! - 4/4

25 = 4! + 4/4
26 = 4sq + 4/.4
27 = 4! + 4 - 4/4
28 = 4sq + 4sq - 4
29 = 4! + 4 + 4/4
30 = (4+4/4)!/4
31 = 4sq + 4sq - 4/4
32 = 4sq + 4sq
33 = 4sq + 4sq + 4/4
34 = 4! + 4/.4
35 = 4! + 44/4
37 = 4!sq / 4sq + 4/4
39 = 4! + 4sq - 4/4
40 = 4! + 4sq
41 = 4! + 4sq + 4/4

95 = (4! * 4sq-4)/4
96 = 4! * 4sq/4
97 = (4! * 4sq+4)/4
109 = (44 -.4)/.4
120 = (4+4/4)!
121 = (44/4)sq

7 = (4!+4)/4
60 = 4!/.4

14 = 4sq - (4+4)/4
22 = 4! - (4+4)/4
43 = 44 - 4/4
45 = 44 + 4/4
47 = 4! + 4! - 4/4
49 = 4! + 4! + 4/4
52 = 4! + 4! + 4

56 = 4*4sq-4-4
60 = 4*4sq-4
63 = 4*4sq-4/4
65= 4*4sq+4/4
81 = ((4-4/4)sq)sq

hmm. The way I remember this was that you had to use exactly 4 4's

I don't think squaring should be included, I can't logically fathom why that's allowed. Perhaps they mean square roots?

Also, is (.4...) or .4 with an overbar allowed? (4/.4..) is a nice way to get 9 :)

Has anyone ever looked at a systematic way of enumerating all possible answers?

[QUOTE=Numbers]Did I miss something ?[/QUOTE]You are miss[i]ing[/i] something, and that is the list of numbers between 1 and 160 for which I could and couldn't find solutions. But I want you to try to discover those for yourself, as I may have missed solutions for some of them. :wink:

[quote=TravisT]hmm. The way I remember this was that you had to use exactly 4 4's[/Quote]That's the way it was in the book, but I explained earlier that since it's easy to add 4's, I'll go for the smallest number of 4's possible.

[quote=TravisT]I don't think squaring should be included, I can't logically fathom why that's allowed. Perhaps they mean square roots?

Also, is (.4...) or .4 with an overbar allowed? (4/.4..) is a nice way to get 9 :)

Has anyone ever looked at a systematic way of enumerating all possible answers?[/quote]It's this way because it's from a book for doing math on a calculator that has an x^2 button, and because it's more challenging this way. I found several web pages using overbars, square roots, and of course the general method of square roots and logs mentioned in the "Generating 2005" puzzle. I like the description from [url=http://www.math.toronto.edu/mathnet/questionCorner/fourfours.html]this page[/url]:
[quote]In summary, this puzzle depends entirely on what rules you choose. It has more to do with manipulating the symbolism in clever ways than on any mathematical truths, and if mathematical notation had evolved differently, the outcome of the puzzle would be quite different too.[/quote]This particular set of rules seems to allow a wide range of solutions, but makes them harder to find than some other rule sets.

P.S. There's a better solution for 12 (and thus 3).

Edit: I guessed 160 as a rough upper limit on the contiguous integers from 1 that have four 4's solutions. We can go higher if anyone wants.

12 = 4sq-4
40 = 4sq/.4

Thru 75, I'm having some trouble computing the following:

38, 42, 55, 57, 58, 62, 67 and 75

[QUOTE=grandpascorpion]Thru 75, I'm having some trouble computing the following:

38, 42, 55, 57, 58, 62, 67 and 75[/QUOTE]

38 = 4! + 4/.4 + 4

42 = 4! + 4! - 4!/4

58 = 4! + 4! + 4/.4

I wonder how difficult it would be to write a program to evaluate all possible expressions (given the limitations) and output the simplest solution for each, i.e. the one that uses the fewest 4's.

Spoilers at: [url]http://www.cut-the-knot.org/arithmetic/funny/4_4.shtml[/url]

Ah but many of those solutions use the square root sign (which is verboten under Ken's rules).

Here are some others without resorting to the link I produced above. If they are the same, it is coincidental. There are multiple solutions to many of these. Many of the ones below follow easily from other solutions.

59 = 4!/.4 - 4/4
61 = 4!/.4 + 4/4
66 = 4!/.4 + 4!/4
68 = 4!/.4 + 4 + 4
68 = 4! + 4! + 4! - 4
70 = (4 + 4)!/(4! * 4!)
76 = 4! + 4! + 4! + 4
82 = 4!/.4 + 4! - 4
86 = 4!/.4 + 4!
90 = 4!/.4 + 4! + 4
92 = 4! * 4 - 4
95 = 4! * 4 - 4/4
97 = 4! * 4 + 4/4
101 = (4/.4)sq + 4/4
104 = (4/.4)sq + 4
108 = (4/.4)sq + 4 + 4
116 = (4+4/4)! - 4
117 = (44/4)sq - 4
124 = (4+4/4)! + 4
125 = (44/4)sq + 4
128 = 4!sq/4 - 4sq
132 = 4!sq/4 - 4sq + 4
136 = 4!sq/4 - 4 - 4
140 = 4!sq/4 - 4
143 = 4!sq/4 - 4/4
144 = 4!sq/4 + 4/4

Here is the current list to save everyone from scrolling. It contains duplicates as those might provide ideas for expanding the table.

0 = 4-4
1 = 4/4
2 = (4+4)/4
3 = (4+4+4)/4
4 = 4
5 = 4 + 4/4
6 = 4! / 4
7 = (4!+4)/4
7 = 4 + 4 - 4/4
8 = 4 + 4
9 = 4 + 4 + 4/4
10 = 4 / .4
11 = 44 / 4
12 = 4 + 4 + 4
12 = 4sq-4
13 = 4sq -4 + 4/4
14 = 4sq - (4+4)/4
15 = 4sq - 4/4
16 = 4sq
17 = 4sq + 4/4
18 = 4/.4 + 4 + 4
19 = 4! - 4 - 4/4
20 = 4! - 4
21 = 4! -4 + 4/4
22 = 4! - (4+4)/4
23 = 4! - 4/4
24 = 4!
25 = 4! + 4/4
26 = 4sq + 4/.4
27 = 4! + 4 - 4/4
28 = 4sq + 4sq - 4
29 = 4! + 4 + 4/4
30 = (4+4/4)!/4
31 = 4sq + 4sq - 4/4
32 = 4sq + 4sq
33 = 4sq + 4sq + 4/4
34 = 4! + 4/.4
35 = 4! + 44/4
36 = 4!sq / 4sq
37 = 4!sq / 4sq + 4/4
38 = 4! + 4/.4 + 4
39 = 4! + 4sq - 4/4
40 = 4! + 4sq
40 = 4sq/.4
41 = 4! + 4sq + 4/4
42 = 4! + 4! - 4!/4
43 = 44 - 4/4
44 = 44
45 = 44 + 4/4
47 = 4! + 4! - 4/4
48 = 4! + 4!
49 = 4! + 4! + 4/4
52 = 4! + 4! + 4
56 = 4*4sq-4-4
58 = 4! + 4! + 4/.4
59 = 4!/.4 - 4/4
60 = 4!/.4
60 = 4*4sq-4
61 = 4!/.4 + 4/4
63 = 4*4sq-4/4
64 = 4sq * 4
65 = 4*4sq+4/4
66 = 4!/.4 + 4!/4
68 = 4! + 4! + 4! - 4
68 = 4!/.4 + 4 + 4
70 = (4 + 4)!/(4! * 4!)
72 = 4! * 4 - 4!
76 = 4! + 4! + 4! + 4
80 = 4sq * 4 + 4sq
81 = ((4-4/4)sq)sq
82 = 4!/.4 + 4! - 4
86 = 4!/.4 + 4!
90 = 4!/.4 + 4! + 4
92 = 4! * 4 - 4
95 = (4! * 4sq-4)/4
95 = 4! * 4 - 4/4
96 = 4! * 4
96 = 4! * 4sq/4
97 = (4! * 4sq+4)/4
97 = 4! * 4 + 4/4
100 = ( 4 / .4 )sq
101 = (4/.4)sq + 4/4
104 = (4/.4)sq + 4
108 = (4/.4)sq + 4 + 4
109 = (44 -.4)/.4
110 = 44 / .4
111 = 44.4 / .4
116 = (4+4/4)! - 4
117 = (44/4)sq - 4
120 = (4+4/4)!
121 = (44/4)sq
124 = (4+4/4)! + 4
125 = (44/4)sq + 4
128 = 4!sq/4 - 4sq
128 = 4sq * 4 + 4sq * 4
132 = 4!sq/4 - 4sq + 4
136 = 4!sq/4 - 4 - 4
140 = 4!sq/4 - 4
143 = 4!sq/4 - 4/4
144 = 4!sq / 4
144 = 4!sq/4 + 4/4
256 = 4sq * 4sq

And some more:

3 = 4 - 4/4
50 = 4sqsq - 4/.4 - 4
53 = 4sqsq - 44/4
54 = 4sqsq - 4/.4
54 = 4!/.4 - 4!/4
54 = 4! + 4! + 4!/4
55 = 4sqsq - (4 - 4/4)sq
57 = 4sqsq - (4!+4)/4
62 = 4sqsq - (4+4)/4
64 = 4sqsq
67 = 4sqsq + 4 - 4/4
69 = 4sqsq + 4 + 4/4
71 = 4sqsq + (4!+4)/4
73 = 4sqsq + (4 - 4/4)sq
74 = 4sqsq + 4/.4
75 = 4sqsq + 44/4
78 = 4sqsq + 4/.4 + 4
79 = 4sqsq + 4sq - 4/4
81 = 4sqsq + 4sq + 4/4
84 = 4sq * 4 + 4sq + 4
87 = 4sqsq + 4! - 4/4
89 = 4sqsq + 4! + 4/4
94 = (4/.4)sq - 4!/4
99 = (4/.4)sq - 4/4
102 = 4! * 4 + 4!/4
106 = (4/.4)sq + 4!/4
112 = 4! * 4 + 4sq
114 = 44/.4 + 4

S'more
 

46 = (4sq+4)/.4-4
50 = (4sq+4)/.4
51 = (4sq+4+.4)/.4
53 = ((4!+4)/4)sq+4
54 = (4sq+4)/.4+4
69 = (4!+4-.4)/.4
71 = (4!+4+.4)/.4
74 = (4!+4)/.4+4
84 = 4*(4sq+4)+4
94 = ( 4 / .4 )sq-4!/4
99 = ( 4 / .4 )sq-4/4
106 = ( 4 / .4 )sq+4!/4
112 = 4*(4!+4)

edit : sorry, there's redundancies

How did I miss this one:

88 = 4sqsq + 4!

and

77 = (4 - 4/4)sqsq - 4
85 = (4! + 4/.4)/.4 (this one is a cheat from the link I provided)
85 = (4 - 4/4)sqsq + 4 (this one is mine)
98 = 4sqsq + 4! + 4/.4
105 = (4 - 4/4)sqsq + 4!

That leaves 83, 91, 93, and 107 without solutions under 110.

0 = 4-4
1 = 4/4
2 = (4+4)/4
3 = 4-4/4
4 = 4
5 = 4+4/4
6 = 4+(4+4)/4 = 4/.4-4 = 4!/4
7 = 4+4-4/4 = (4!+4)/4
8 = 4+4
9 = (4-.4)/.4
10 = 4/.4
11 = 44/4
12 = 4^2-4
13 = 4^2-4+4/4 = (4!+4!+4)/4
14 = 4^2-(4+4)/4
15 = 4^2-4/4
16 = 4^2
17 = 4^2+4/4
18 = 4^2+(4+4)/4
19 = (4+4-.4)/.4
20 = (4+4)/.4
21 = (4+4.4)/.4
22 = 4!-(4+4)/4
23 = 4!-4/4
24 = 4!
25 = 4!+4/4
26 = 4!+(4+4)/4
27 = 4!+4-4/4
28 = 4!+4
29 = 4!+4+4/4
30 = (4+4/4)!/4 = (4+4+4)/.4
31 = 4^2+4^2-4/4
32 = 4^2+4^2
33 = 4^2+4^2+4/4
34 = 4! + 4/.4
35 = 4! + 44/4
36 = 4*(4-.4)/.4 = (4!/4)^2
37 = (4!/4)^2 + 4/4
38 = (4^2-.4-.4)/.4
39 = (4^2-.4)/.4
40 = 4^2/.4
41 = (4^2+.4)/.4
42 = (4^2+.4+.4)/.4
43 = 44-4/4
44 = 44
45 = 44+4/4

47 = 4!+4!-4/4
48 = ((4!+4)^2-4^2)/4^2 = 4!*(4+4)/4 = 4*(4^2-4) = 44+4 = 4!+4!
49 = (4+4-4/4)^2 = 4!+4!+4/4 = ((4!+4)/4)^2
50 = ((4!+4)^2+4^2)/4^2 = (4!-4)/.4
51 = (4!-4+.4)/.4
52 = 44+4+4 = 4!+4!+4
53 = ((4!+4)/4)^2+4

55 = (4!-.4)/.4-4
56 = 4!+4!+4+4 = 4!/.4-4
57 = (4!+.4)/.4-4
58 = (4!-.4-.4)/.4
59 = (4!-.4)/.4
60 = 4!/.4
61 = (4!+.4)/.4
62 = (4!+.4+.4)/.4
63 = 4*4^2-4/4
64 = 4*4^2 = (4+4)^2
65 = 4*4^2+4/4 = ((4!+4)/4)^2+4^2

68 = 4!+44
69 = (4!+4-.4)/.4
70 = (4!+4)/.4
71 = (4!+4.4)/.4
72 = 4!+4!+4!

74 = (4!+4)/.4+4

76 = 4!+4!+4!+4

80 = (4!+4+4)/.4 = (4^2-4)^2-(4+4)^2 = 4*(4!-4)
81 = ((4-.4)/.4)^2

84 = 4*(4!-4)+4 = 4!/.4+4!
85 = ((4-.4)/4)^2+4 = (4!+.4)/.4+4!

88 = 4!+4!+4!+4^2

92 = (4/.4)^2-4-4 = 4*4!-4

95 = 4!*4-4/4
96 = 4!*4
97 = 4!*4+4/4

99 = (4/.4)^2-4/4
100 = 4*4!+4 = (4/.4)^2
101 = (4/.4)^2+4/4

104 = (4/.4)^2+4
105 = (44/4)^2-4^2
106 = 44/.4-4

108 = (4/.4)^2+4+4
109 = (44-.4)/.4
110 = 44/.4
111 = 444/4
112 = (4^2-4)^2-4^2-4^2

114 = 44/.4+4

116 = (4/.4)^2+4^2
117 = (44/4)^2-4

119 = (4!+4!-.4)/.4
120 = (4+4/4)! = (4!+4!)/.4
121 = (44/4)^2

124 = (4^2-4)^2-4^2-4
125 = (44/4)^2+4 = (4!-4)/(.4)^2

128 = (4^2-4)^2-4^2

132 = (4/.4)^2+4^2+4^2 = (4^2-4)^2-4^2+4

136 = (4^2-4)^2-4-4
137 = (44/4)^2+4^2
138 = (4!^2-4!)/4

140 = (4^2-4)^2-4

142 = (4!^2-4-4)/4
143 = (4^2-4)^2-4/4 = (4!^2-4)/4
144 = (4^2-4)^2 = 4!^2/4
145 = (4^2-4)^2+4/4 = (4!^2+4)/4
146 = (4!^2+4+4)/4

148 = (4^2-4)^2+4
149 = (4!-(.4)^2)/(.4)^2
150 = 4!/(.4)^2
151 = (4!+(.4)^2)/(.4)^2
152 = (4^2-4)^2+4+4 = 4*44-4!
153 = 4+(4!-(.4)^2)/(.4)^2
154 = 4+4!/(.4)^2
155 = 4+(4!+(.4)^2)/(.4)^2

160 = (4^2-4)^2+4^2

Rogue: Good job! Your 38 beats what I had, which was (4!sq+4sq+4sq)/4sq. I haven't had a chance to look at the rest of the solutions posted since then.

Someone could probably modify the code I wrote to solve the "Generating 2005" puzzle (which is attached to a post in there) to solve this. I haven't looked at trying yet.

P.S. That site is actually the worst of three sites with spoilers I found with Google. That site also uses the illegal floor (integer part) function.

rogue: what do you mean by ()sqsq? If I read it as (()^2)^2, 4sqsq + 4! = 65560.
BTW: 82 still lacks a solution, as 4!/.4 + 4! - 4 = 80 /= 82

Benjamin

Rogue ,

There's a problem with many of your solutions:

4sqsq isn't 64, it's 256. But where you use that with three 4's or less, you can sub your 4sqsq with 4*4sq.

62 = (4sqsq - 4 -4)/4

89 = ((4!/4)sq-.4)/.4
90 = (4!/4)sq/.4
91 = ((4!/4)sq+.4)/.4

Oops. It must have been that beer I had this evening. I don't know what I was thinking.

One little trick:

If you can't find a solution, see if 256-x is possible

113 = 4sqsq - (4!sq-4)/4
118 = 4sqsq - (4!sq-4!)/4
135 = 4sqsq - (44/4)sq
156 = 4sqsq - (4/.4)sq

That's as far as I checked.

Just recognised that you can save a 4 in my solution for 63/65 addopting this: 63 = (4^2^2-4)/4 giving 67 = (4^2^2-4)/4+4

46 = (4!-4)/.4-4
54 = 4!/4+4!+4!
66 = (4!+4)/.4-4
130 = (4!+4!+4)/.4

smallest # w/o representation: 73

Glancing over the record, I saw that 154 has a 3 4's representation, so:
158 = 4!/(.4)^2+4+4

73 = (4sqsq+(4!/4)sq)/4
127 = (4!sq-4)/4-4sq
139 = (4!sq-4)/4-4
147 = (4!sq-4)/4+4
159 = (4!sq-4)/4+4sq

Where is "75" ?

From above:
77 = (4 - 4/4)sqsq - 4
94 = (4/.4)sq - 4!/4
105 = (44/4)^2-4^2

126 = (4!sq+4!)/4-4!
129 = (4!sq+4)/4-4sq
134 = (4!sq+4!)/4 - 4sq
137 = (4!sq-4!-4)/4
141 = (4!sq-4sq+4)/4

75 = (4sqsq+44)/4
86 = 4!/(.4sq)-4*4sq
122= 4!/(.4sq)-4!-4

133 = (4!-.4sq)/(.4sq)-4sq

pi = (-4/(4+4))!sq

:D

[QUOTE="Wacky"] Where is "75" ?[/QUOTE]Oops! forgot to post:
75 = (4!+4!/4)/.4

So, we're down to 13.

Ken,

Thanks for the "puzzle". Perhaps you will now disclose which of the remaining numbers are those that you have not yet solved.

As for the "Scoreboard", I haven't found a way to get a message to the top of the thread.

So, for the next thread for this kind of puzzle, let me suggest that the original author set up the top of the thread by posting two or three messages.

(1) The statement of the problem (to be edited if clarification is required)
(2) A placeholder for the Scoreboard
(3) The original statement of the problem (and author's comments)

That way, we can have the "executive overview" of the puzzle and scoreboard at the head and the rest of the thread can develop in a normal fashion.

Thanks, GrandpaScorpion! That subtract-from-256 trick solved one I hadn't gotten (113), and helped me get two others. So now there's only one less than 160 that I haven't gotten. :bounce: I'd rather not disclose it just yet, as it might discourage you from finding a solution for it!

Trying numbers up around 256, it looks to me like it might be fruitful to expand the range up to 300, though I wouldn't expect to find solutions for all numbers in that range.

Wacky, can't you just edit your first message (the 3rd in the list) this time?

I think the solution to 73 is incorrect. Try this:

73 = ((4! + 4)/4)sq + 4!

And two more:

79 = (4sqsq - 4)/4 + 4sq
87 = (4sqsq - 4)/4 + 4!

73 = (4sqsq+(4!/4)sq)/4 is valid
= (256 + 6^2)/4 = 73

Is it me or did some solutions disappear (like 79 and 87 earlier)?

And Ken, glad that helped. I'm game for increasing finding solutions thru 300.

[QUOTE=grandpascorpion]73 = (4sqsq+(4!/4)sq)/4 is valid
= (256 + 6^2)/4 = 73[/QUOTE]

You are correct and I noticed that right after I posted.

Thanks

Here is the current list. I have removed duplicates and solutions which use more fours than other solutions for the same number.

0 = 4 - 4
1 = 4/4
2 = (4 + 4)/4
3 = 4 - 4/4
4 = 4
5 = 4 + 4/4
6 = 4!/4
7 = (4! + 4)/4
8 = 4 + 4
9 = (4 - .4)/.4
10 = 4 / .4
10 = 4/.4
11 = 44 / 4
12 = 4sq - 4
13 = 4sq - 4 + 4/4
14 = 4sq - (4 + 4)/4
15 = 4sq - 4/4
16 = 4sq
17 = 4sq + 4/4
18 = 4/.4 + 4 + 4
18 = 4sq + (4 + 4)/4
19 = (4 + 4 - .4)/.4
19 = 4! - 4 - 4/4
20 = 4! - 4
21 = (4 + 4.4)/.4
21 = 4! - 4 + 4/4
22 = 4! - (4 + 4)/4
23 = 4! - 4/4
24 = 4!
25 = 4! + 4/4
26 = 4sq + 4/.4
27 = 4! + 4 - 4/4
28 = 4! + 4
29 = 4! + 4 + 4/4
30 = (4 + 4/4)!/4
31 = 4sq + 4sq - 4/4
32 = 4sq + 4sq
33 = 4sq + 4sq + 4/4
34 = 4! + 4/.4
35 = 4! + 44/4
36 = (4!/4)sq
36 = 4!sq / 4sq
37 = (4!/4)sq + 4/4
37 = 4!sq / 4sq + 4/4
38 = (4sq - .4 - .4)/.4
38 = 4! + 4/.4 + 4
39 = (4sq - .4)/.4
40 = 4! + 4sq
40 = 4sq/.4
41 = (4sq + .4)/.4
41 = 4! + 4sq + 4/4
42 = (4sq + .4 + .4)/.4
42 = 4! + 4! - 4!/4
43 = 44 - 4/4
44 = 44
45 = 44 + 4/4
46 = (4! - 4)/.4 - 4
46 = (4sq + 4)/.4 - 4
47 = 4! + 4! - 4/4
48 = 4! + 4!
49 = ((4! + 4)/4)sq
50 = (4! - 4)/.4
50 = (4sq + 4)/.4
51 = (4! - 4 + .4)/.4
51 = (4sq + 4 + .4)/.4
52 = 4! + 4! + 4
52 = 44 + 4 + 4
53 = ((4! + 4)/4)sq + 4
54 = (4sq + 4)/.4 + 4
54 = 4! + 4! + 4!/4
54 = 4!/.4 - 4!/4
54 = 4!/4 + 4! + 4!
55 = (4! - .4)/.4 - 4
56 = 4!/.4 - 4
57 = (4! + .4)/.4 - 4
58 = (4! - .4 - .4)/.4
58 = 4! + 4! + 4/.4
59 = (4! - .4)/.4
60 = 4!/.4
61 = (4! + .4)/.4
61 = 4!/.4 + 4/4
62 = (4! + .4 + .4)/.4
63 = 4sq * 4 - 4/4
64 = (4 + 4)sq
64 = 4sq * 4
65 = 4sq * 4 + 4/4
66 = (4! + 4)/.4 - 4
66 = 4!/.4 + 4!/4
67 = (4sqsq - 4)/4 + 4
68 = 4! + 44
68 = 4sq * 4 + 4
69 = (4! + 4 - .4)/.4
70 = (4! + 4)/.4
71 = (4! + 4.4)/.4
72 = 4! * 4 - 4!
72 = 4! + 4! + 4!
73 = ((4! + 4)/4)sq + 4!
74 = (4! + 4)/.4 + 4
75 = (4! + 4!/4)/.4
76 = 4! + 4! + 4! + 4
77 = (4 - 4/4)sqsq - 4
79 = (4sqsq - 4)/4 + 4sq
80 = 4*(4! - 4)
80 = 4sq * 4 + 4sq
81 = ((4 - .4)/.4)sq
81 = ((4 - 4/4)sq)sq
84 = 4!/.4 + 4!
84 = 4*(4sq + 4) + 4
85 = (4 - 4/4)sqsq + 4
85 = (4! + .4)/.4 + 4!
85 = (4! + 4/.4)/.4
86 = 4!/(.4sq) - 4sq * 4
87 = (4sqsq - 4)/4 + 4!
88 = 4! + 4! + 4! + 4sq
89 = ((4!/4)sq - .4)/.4
90 = (4!/4)sq/.4
91 = ((4!/4)sq + .4)/.4
92 = 4! * 4 - 4
94 = (4 / .4)sq - 4!/4
94 = (4/.4)sq - 4!/4
95 = (4! * 4sq - 4)/4
95 = 4! * 4 - 4/4
96 = 4! * 4
97 = (4! * 4sq + 4)/4
97 = 4! * 4 + 4/4
99 = (4 / .4)sq - 4/4
99 = (4/.4)sq - 4/4
100 = (4/.4)sq
101 = (4/.4)sq + 4/4
102 = 4! * 4 + 4!/4
104 = (4/.4)sq + 4
105 = (4 - 4/4)sqsq + 4!
105 = (44/4)sq - 4sq
106 = (4/.4)sq + 4!/4
106 = 44/.4 - 4
108 = (4/.4)sq + 4 + 4
109 = (44 - .4)/.4
110 = 44/.4
111 = 444/4
112 = 4 * (4! + 4)
112 = 4! * 4 + 4sq
113 = 4sqsq - (4!sq - 4)/4
114 = 44/.4 + 4
116 = (4/.4)sq + 4sq
117 = (44/4)sq - 4
118 = 4sqsq - (4!sq - 4!)/4
119 = (4! + 4! - .4)/.4
120 = (4 + 4/4)!
120 = (4! + 4!)/.4
121 = (44/4)sq
122 = 4!/(.4sq)-4! - 4
124 = (4 + 4/4)! + 4
124 = (4sq - 4)sq - 4sq - 4
125 = (44/4)sq + 4
125 = (4! - 4)/(.4)sq
126 = (4!sq + 4!)/4 - 4!
127 = (4!sq - 4)/4 - 4sq
128 = (4sq - 4)sq - 4sq
128 = 4!sq/4 - 4sq
128 = 4sq * 4 + 4sq * 4
129 = (4!sq + 4)/4 - 4sq
130 = (4! + 4! + 4)/.4
132 = (4/.4)sq + 4sq + 4sq
132 = (4sq - 4)sq - 4sq + 4
132 = 4!sq/4 - 4sq + 4
133 = (4! - .4sq)/(.4sq)-4sq
134 = (4!sq + 4!)/4 - 4sq
135 = 4sqsq - (44/4)sq
136 = (4sq - 4)sq - 4 - 4
136 = 4!sq/4 - 4 - 4
137 = (4!sq - 4! - 4)/4
137 = (44/4)sq + 4sq
138 = (4!sq - 4!)/4
139 = (4!sq - 4)/4 - 4
140 = (4sq - 4)sq - 4
140 = 4!sq/4 - 4
141 = (4!sq - 4sq + 4)/4
142 = (4!sq - 4 - 4)/4
143 = (4!sq - 4)/4
143 = 4!sq/4 - 4/4
144 = (4sq - 4)sq
144 = 4!sq/4
144 = 4!sq/4 + 4/4
145 = (4!sq + 4)/4
146 = (4!sq + 4 + 4)/4
147 = (4!sq - 4)/4 + 4
148 = (4sq - 4)sq + 4
149 = (4!-(.4)sq)/(.4)sq
150 = 4!/(.4)sq
151 = (4!+(.4)sq)/(.4)sq
152 = (4sq - 4)sq + 4 + 4
152 = 4 * 44 - 4!
153 = 4+(4!-(.4)sq)/(.4)sq
154 = 4 + 4!/(.4)sq
155 = 4+(4!+(.4)sq)/(.4)sq
156 = 4sqsq - (4/.4)sq
158 = 4!/(.4)sq + 4 + 4
159 = (4!sq - 4)/4 + 4sq
160 = (4sq - 4)sq + 4sq
256 = 4sqsq

According to [url="http://ourworld.compuserve.com/homepages/DavidandPenny/200Up.htm"]http://ourworld.compuserve.com/homepages/DavidandPenny/200Up.htm[/url] the remaining 4-4's-representations up to 160 must have at least one squaring operation.

83 = (4! -.4)/.4 + 4!


Added
162 = 4!/(.4)sq + 4sq - 4
170 = 4!/(.4)sq + 4sq + 4
178 = 4!/(.4)sq + 4! + 4
182 = 4!/(.4)sq + 4sq + 4sq
186 = 4!/(.4)sq
194 = 4!/(.4)sq + 44

83 = (4!-.4)/.4+4!

165 = (4!-.4[sup]sq[/sup])/(.4[sup]sq[/sup])+4[sup]sq[/sup]

167 = (4![sup]sq[/sup]-4)/4+4!
168 = 4![sup]sq[/sup]/4+4!
169 = (4![sup]sq[/sup]+4)/4+4!

173 = (4!-.4[sup]sq[/sup])/(.4[sup]sq[/sup])+4!

I thought someone earlier posted: 192 = 4!*4+4!*4

And we've been using: 576 = 4!sq

174 = (4!)/(.4sq)+4!
175 = (4!+.4sq)/(.4sq)+4!

161= 159 + 2, 164 = 4 * 41
 

161 = (4!sq + 4)/4 + 4sq
164 = 4*(4sq + .4)/.4

Solutions with fewer 4's than on the last summary:

14 = 4/.4+4
14 = 4! -4/.4

22 = 4sq+4!/4

30 = 4!+4!/4

63 = (4[sup]sqsq[/sup]-4)/4

65 = (4[sup]sqsq[/sup]+4)/4

76 = (4/.4)[sup]sq[/sup]-4!

124 = (4/.4)[sup]sq[/sup]+4!

126 = 4!/(.4[sup]sq[/sup])-4!

143 = (4![sup]sq[/sup]-4)/4

146 = 4!/(.4[sup]sq[/sup])-4

and a typo correction:

144 = 4!sq/4 + 4/4 should have been
14[b]5[/b] = 4!sq/4 + 4/4

Earlier, while under the influence of beer :-) , rogue posted several solutions in which 4[sup]sqsq[/sup] was erroneously evaluated to equal 64 instead of 256.

[quote="rogue"]50 = 4sqsq - 4/.4 - 4
53 = 4sqsq - 44/4
54 = 4sqsq - 4/.4
55 = 4sqsq - (4 - 4/4)sq
57 = 4sqsq - (4!+4)/4
62 = 4sqsq - (4+4)/4
64 = 4sqsq
67 = 4sqsq + 4 - 4/4
69 = 4sqsq + 4 + 4/4
71 = 4sqsq + (4!+4)/4
73 = 4sqsq + (4 - 4/4)sq
74 = 4sqsq + 4/.4
75 = 4sqsq + 44/4
78 = 4sqsq + 4/.4 + 4
79 = 4sqsq + 4sq - 4/4
81 = 4sqsq + 4sq + 4/4
87 = 4sqsq + 4! - 4/4
88 = 4sqsq + 4!
89 = 4sqsq + 4! + 4/4
98 = 4sqsq + 4! + 4/.4
[/quote]

All that work by rogue can be salvaged by simply adding 192 to the left side of each of those equations:

242 = 4sqsq-4/.4-4
245 = 4sqsq-44/4
246 = 4sqsq-4/.4
247 = 4sqsq-(4-4/4)sq
249 = 4sqsq-(4!+4)/4
254 = 4sqsq-(4+4)/4
256 = 4sqsq
259 = 4sqsq+4-4/4
261 = 4sqsq+4+4/4
263 = 4sqsq+(4!+4)/4
265 = 4sqsq+(4-4/4)sq
266 = 4sqsq+4/.4
267 = 4sqsq+44/4
270 = 4sqsq+4/.4+4
271 = 4sqsq+4sq-4/4
273 = 4sqsq+4sq+4/4
279 = 4sqsq+4!-4/4
280 = 4sqsq+4!
281 = 4sqsq+4!+4/4
290 = 4sqsq+4!+4/.4

[QUOTE=Wacky][b]These still need a solution[/b]
78
93
107
131
157

161-255[/QUOTE]

(0.4[sup]sq[/sup]+(4-.4)[sup]sq[/sup])/0.4[sup]sq[/sup]=82
4+(4[sup]sq[/sup]-.4[sup]sq[/sup])/.4[sup]sq[/sup]=103
4[sup]sq[/sup]+(4[sup]sq[/sup]-.4[sup]sq[/sup])/.4[sup]sq[/sup]=115
4!+(4[sup]sq[/sup]-.4[sup]sq[/sup])/.4[sup]sq[/sup]=123

And, impressingly enough:

[QUOTE=S80780]According to [url="http://ourworld.compuserve.com/homepages/DavidandPenny/200Up.htm"]http://ourworld.compuserve.com/homepages/DavidandPenny/200Up.htm[/url] the remaining 4-4's-representations up to 160 must have at least one squaring operation.[/QUOTE]

4*(.4+4!)+.4=98 :showoff:

... and fetofs's solutions lead to:

99 = (4[sup]sq[/sup]-.4[sup]sq[/sup])/.4[sup]sq[/sup]

101 = (4[sup]sq[/sup]+.4[sup]sq[/sup])/.4[sup]sq[/sup]

... which lead to:

157 = 4[sup]sqsq[/sup]-(4[sup]sq[/sup]-.4[sup]sq[/sup])/.4[sup]sq[/sup]

[QUOTE=fetofs][QUOTE=Wacky]
93
107
131

163
166
171
172

176
(and more)[/QUOTE][/QUOTE]

78=4[sup]sq[sup]sq[/sup][/sup]*.4-(.4+4!)

Working on 93.

:banana:

Well, I guess it's time to post solutions for a few you hadn't gotten earlier (I don't know if you've solved them yet:

78 = 4!*4!sq/4sqsq+4!
82 = (4sqsq-4!)/4+4!
83 = (4sq-.4sq)/.4sq-4sq
98 = (4sq-.4sq-.4sq)/.4sq
103= (4sq-.4sq)/.4sq+4
107= 4sqsq-(4!-.4sq)/.4sq
115= (4sq-.4sq)/.4sq+4sq
123= (4sq-.4sq)/.4sq+4!
131= 4sqsq-(4sq+4)/.4sq
157= 4sqsq-(4sq-.4sq)/.4sq

93 is the only one under 160 I haven't found. Between 160 and 300, I couldn't find the following 25 (yet):

163
177
181
183
185
187
189
202
203
205
206
211
213
214
217
218
223
237
243
269
283
287
294
295
298

Also, I found a copy of the problem that's apparently from an earlier edition of my book on [url=http://www.geocities.com/TimesSquare/Arcade/7810/44sfaq.htm]this page[/url]. It's almost the same as mine, but his is copyright 1976 and mine is 1980. Basically, that sentence allowing 4! and repeating decimals is gone in mine, and the ! key is allowed generally.

Do you believe that there is a solution for all of these?

[QUOTE=Ken_g6] Basically, that sentence allowing 4! and repeating decimals is gone in mine, and the ! key is allowed generally.[/QUOTE]

I think repeated decimals should be allowed, or else the book wouldn't claim solutions up to 130 (while there's no solution for 93 I have found on the net).

((4 - .4)[sup]sq[/sup]+4[sup]sq[/sup])/.4[sup]sq[/sup]=181
((4+4!)[sup]sq[/sup]+4!)/4=202

176 = 4 * 44
172 = 4 * 44 - 4
180 = 4 * 44 + 4
200 = 4 * 44 + 4!
250 = 4sqsq - 4!/4

[QUOTE=rogue]Do you believe that there is a solution for all of these?[/QUOTE]
No. I rather doubt that many, if any, can be solved. But I thought that about 113, too.

fetofs: That's the first time I've seen a solution that produces, then subtracts away .4. I'll see if I can find any other uses for that.

And I never saw anything about repeating decimals until I searched the internet. But the book could be wrong.

A few solutions in 3/4 time:

54 = 4!*4![sup]sq[/sup]/4[sup]sq[sup]sq[/sup][/sup]

58 = (4[sup]sq[sup]sq[/sup][/sup]-4!)/4

102 = 4[sup]sq[sup]sq[/sup][/sup]*.4-.4

166 = 4!/.4[sup]sq[/sup]+4[sup]sq[/sup]

196 = (4!+4)[sup]sq[/sup]/4

214=(4+4)[sup]sq[/sup]+4!/.4[sup]sq[/sup]
294=((4/.4)[sup]sq[/sup]-4[sup]sq[/sup])[sup]sq[/sup]/4!
269=(.4+4[sup]sq[/sup])[sup]sq[/sup]+.4[sup]sq[/sup]/4

15^2 based answers and more
 

249 = (4!/.4(4))sq - 4!
209 = (4!/.4(4))sq - 4sq
221 = (4!/.4(4))sq -4
225 = (4!/.4(4))sq
229 = (4!/.4(4))sq + 4
241 = (4!/.4(4))sq + 4sq
249 = (4!/.4(4))sq + 4!

204 = (4!sq+4sqsq)/4-4
207 = (4!sq+4sqsq-4)/4
208 = (4!sq+4sqsq)/4
209 = (4!sq+4sqsq+4)/4
210 = (4!/.4+4!)/.4
212 = (4!sq+4sqsq)/4+4
216 = 4sqsq-4!-4sq
220 = 4sqsq -4sq -4sq-4

Most of the solutions near 256 are easy just use 256-x where x uses 3 4's.

Also Found a pretty one for 180

4! + 4! + 4!
-----------
.4

[QUOTE=fetofs]And, impressingly enough: [snip]
4*(.4+4!)+.4=98 :showoff:[/QUOTE]
:redface:Sorry for the misinformation! Should've let the programme run over the complete range and not entering each unknown candidate seperately. You get a slipping one too easily.

Benjamin

I dub fetofs the Four-Four King. :bow:

Here are the ones Wacky listed as missing, that I've found. It takes a while to pull them out of the cross-linked Excel file I make them in. Groupings show how I got some of the numbers.

171 = ((4-4/4)[sup]sq[/sup]+4)[sup]sq[/sup]

175 = (4!+4)/.4[sup]sq[/sup]
179 = (4!+4)/.4[sup]sq[/sup]+4

144 = 4![sup]sq[/sup]/4
168 = 4![sup]sq[/sup]/4+4!
184 = 4![sup]sq[/sup]/4+4!+4[sup]sq[/sup]

192 = 4[sup]sq[sup]sq[/sup][/sup]-4[sup]sq[sup]sq[/sup][/sup]/4
188 = 4[sup]sq[sup]sq[/sup][/sup]-4[sup]sq[sup]sq[/sup][/sup]/4-4

150 = 4!/.4[sup]sq[/sup]
174 = 4!/.4[sup]sq[/sup]+4!
190 = 4!/.4[sup]sq[/sup]+4!+4[sup]sq[/sup]


191 = (4!+4)/.4[sup]sq[/sup]+4[sup]sq[/sup]

193 = 4[sup]sq[sup]sq[/sup][/sup]-(4[sup]sq[sup]sq[/sup][/sup]-4)/4

196 = 4[sup]sq[sup]sq[/sup][/sup]-4!/.4
195 = 4[sup]sq[sup]sq[/sup][/sup]-(4!+.4)/.4

197 = 4[sup]sq[sup]sq[/sup][/sup]-(4!-.4)/.4

198 = 4!/.4[sup]sq[/sup]+4!+4!

199 = (4!+4)/.4[sup]sq[/sup]+4!

200 = (4[sup]sq[/sup]+4[sup]sq[/sup])/.4[sup]sq[/sup]
201 = (4[sup]sq[/sup]+4[sup]sq[/sup]+.4[sup]sq[/sup])/.4[sup]sq[/sup]

204 = (4[sup]sq[/sup]+4[sup]sq[/sup])/.4[sup]sq[/sup]+4

207 = 4[sup]sq[sup]sq[/sup][/sup]-4/.4[sup]sq[/sup]-4!

208 = 4[sup]sq[sup]sq[/sup][/sup]-4!-4!

209 = (4!/(4*.4))[sup]sq[/sup]-4[sup]sq[/sup]

100 = (4/.4)[sup]sq[/sup]
84 = (4/.4)[sup]sq[/sup]-4[sup]sq[/sup]
210 = ((4/.4)[sup]sq[/sup]-4[sup]sq[/sup])/.4

212 = 4[sup]sq[sup]sq[/sup][/sup]-44

231 = 4[sup]sq[sup]sq[/sup][/sup]-4/.4[sup]sq[/sup]
215 = 4[sup]sq[sup]sq[/sup][/sup]-4/.4[sup]sq[/sup]-4[sup]sq[/sup]

232 = 4[sup]sq[sup]sq[/sup][/sup]-4!
216 = 4[sup]sq[sup]sq[/sup][/sup]-4!-4[sup]sq[/sup]

219 = 4[sup]sq[sup]sq[/sup][/sup]-(4![sup]sq[/sup]+4[sup]sq[/sup])/4[sup]sq[/sup]

220 = 4[sup]sq[sup]sq[/sup][/sup]-4![sup]sq[/sup]/4[sup]sq[/sup]

221 = (4!/(4*.4))[sup]sq[/sup]-4

222 = 4[sup]sq[sup]sq[/sup][/sup]-4/.4-4!

224 = 4[sup]sq[sup]sq[/sup][/sup]-4[sup]sq[/sup]-4[sup]sq[/sup]

225 = (4!/(4*.4))[sup]sq[/sup]

Classic problem-Four 4's
 

:yawn:
93 seems to have stumped all of you!
Well here it is
93=4sq(4!/4)-(4-4/4)
Mally :coffee:

Sorry, Mally.
The rules allow ONLY four 4's.

I can do 93 with only 5 4's:

4*4sq+4/.4^2+4

But I haven't figured out 4 4's, yet. :sad:

248 = 4sqsq -4 -4
252 = 4sqsq -4
260 = 4sqsq +4
264 = 4sqsq + 4 + 4
268 = 4sqsq + 4 + 4 + 4
255 = 4sqsq -4/4
251 = 4sqsq -4/4 -4
257 = 4sqsq + 4/4
253 = 4sqsq + 4/4 -4
228 = 4sqsq -4! -4

230 = (4*4! -4) / .4
239 = (4*4! -.4) / .4
240 = 4*4! /.4
236 = (4*4!/.4) - 4
244 = (4*4!/.4) + 4

284 = 4sqsq + 4! + 4
276 = 4sqsq + 4! - 4

262 = 4sqsq + 4!/4
258 = 4sqsq +4!/4 -4

[QUOTE=Ken_g6]171 = ((4-4/4)[sup]sq[/sup]+4)[sup]sq[/sup][/quote]Correction: 169 = ((4-4/4)[sup]sq[/sup]+4)[sup]sq[/sup]

:-)

[QUOTE=cheesehead]Correction: 169 = ((4-4/4)[sup]sq[/sup]+4)[sup]sq[/sup]

:-)[/QUOTE]Correct. Sometimes I look at the Excel row number, which is 2 off. :redface:

175 = (4!+4)/.4[sup]sq[/sup]
171 = (4!+4)/.4[sup]sq[/sup]-4

I have an idea for a means to solve some of the remaining values (focusing more on the odd ones).

4 and 24, each require one 4
3, 5, 7, 11, 15, and 17, each require three 4s

Someone could write a program that could do repeated squarings of these numbers until the difference between two of them is 93. It shouldn't be hard to write, but it is possible that the number of squarings would exceed available memory on ones computer.

[QUOTE=rogue]I have an idea for a means to solve some of the remaining values (focusing more on the odd ones).

4 and 24, each require one 4
3, 5, 7, 11, 15, and 17, each require three 4s

Someone could write a program that could do repeated squarings of these numbers until the difference between two of them is 93. It shouldn't be hard to write, but it is possible that the number of squarings would exceed available memory on ones computer.[/QUOTE]
While it is true that 5 requires three 4s, I observe that 25 = 4/(.4sq). This frees up another 4 to be used elsewhere.

Paul

16(!) left
 

Solutions used with the "256-x" trick:

211=4[sup]sq[sup]sq[/sup][/sup]-(4!/4)!/4[sup]sq[/sup]
217=4[sup]sq[sup]sq[/sup][/sup]-(4[sup]sq[/sup]-.4)/.4

Believe it or not,the "subtract from 256" trick worked out on numbers [b]higher[/b] than 256!

295=4[sup]sq[sup]sq[/sup][/sup]-(.4-4[sup]sq[/sup])/.4

Here "256+x" worked:

185=(4-4/.4[sup]sq[/sup])[sup]sq[/sup]-256

EDIT: grandpascorpion really got it right on this one. Another one I found recently:

206=4[sup]sq[sup]sq[/sup][/sup]-(4 + 4)/.4[sup]sq[/sup]

I can't get beyond 2^31 with Excel, but I tried your ideas, and

I got a few more. I may not have found 93, but I independently came up with 185, 211, and 217, which fetofs just posted.

Here are the ones I found earlier that I hadn't listed here yet:
226 = 4[sup]sq[sup]sq[/sup][/sup]-4!-4!/4
227 = 4[sup]sq[sup]sq[/sup][/sup]-4/.4[sup]sq[/sup]-4
233 = 4[sup]sq[sup]sq[/sup][/sup]+4!-4/4
234 = 4[sup]sq[sup]sq[/sup][/sup]-4[sup]sq[/sup]-4!/4
235 = 4[sup]sq[sup]sq[/sup][/sup]-4/.4[sup]sq[/sup]+4
238 = 4[sup]sq[sup]sq[/sup][/sup]-4!+4!/4

If you haven't found 272, you haven't been looking very hard!
272 = 4[sup]sq[sup]sq[/sup][/sup]+4[sup]sq[/sup]
274 = 4[sup]sq[sup]sq[/sup][/sup]+4!-4!/4
275 = 44/.4[sup]sq[/sup]
277 = 4[sup]sq[sup]sq[/sup][/sup]+4/.4[sup]sq[/sup]-4
278 = 4[sup]sq[sup]sq[/sup][/sup]+4[sup]sq[/sup]+4!/4
282 = 4[sup]sq[sup]sq[/sup][/sup]+4[sup]sq[/sup]+4/.4
285 = 4[sup]sq[sup]sq[/sup][/sup]+4+4/.4[sup]sq[/sup]
286 = 4[sup]sq[sup]sq[/sup][/sup]+4!+4!/4
288 = 4[sup]sq[sup]sq[/sup][/sup]+4[sup]sq[/sup]+4[sup]sq[/sup]
289 = (4[sup]sq[/sup]+4/4)[sup]sq[/sup]
291 = 44/.4[sup]sq[/sup]+4[sup]sq[/sup]
292 = 4[sup]sq[sup]sq[/sup][/sup]+4![sup]sq[/sup]/4[sup]sq[/sup]
293 = 4[sup]sq[sup]sq[/sup][/sup]+(4![sup]sq[/sup]+4[sup]sq[/sup])/4[sup]sq[/sup]
296 = 4[sup]sq[sup]sq[/sup][/sup]+4!+4[sup]sq[/sup]
297 = 4[sup]sq[sup]sq[/sup][/sup]+4/.4[sup]sq[/sup]+4[sup]sq[/sup]
299 = 44/.4[sup]sq[/sup]+4!
300 = 4[sup]sq[sup]sq[/sup][/sup]+44

I just found one with the 576-x trick.
[i]576[/i]-x? 576 = 4![sup]sq[/sup]

287 = 4![sup]sq[/sup]-(4[sup]sq[/sup]+4/4)[sup]sq[/sup]

[i]15[/i] left! :bounce:

Tricky...
 

I tried to work with the "576-x" idea. Guess what... I got a pretty tricky solution!

((4!/.4)[sup]sq[/sup]-4![sup]sq[/sup])/4[sup]sq[/sup]=189

[i]14[/i] left!

[QUOTE=Ken_g6]Concatenate (44); also before (4.4) or after (.44) a Decimal.[/QUOTE]
Ken, would the rules allow any of the following concatenations?

.(4!) = .24

(4[sup]sq[/sup])4 = 164

4(4!) = 424

Perhaps we could require an explicit concatenation operator, such as "cat", except when only plain "4"s are concatenated in a decimal number (e.g., 44, 4.4, 444, .44, 4444, 4.444).

.cat4! = .24

4[sup]sq[/sup]cat4 = 164

4cat4! = 424



We've been using .4[sup]sq[/sup] = .16, but in that case (.4)[sup]sq[/sup] = (.)(4[sup]sq[/sup])

[QUOTE=cheesehead]Ken, would the rules allow any of the following concatenations?

.(4!) = .24

(4[sup]sq[/sup])4 = 164

4(4!) = 424

Perhaps we could require an explicit concatenation operator, such as "cat", except when only plain "4"s are concatenated in a decimal number (e.g., 44, 4.4, 444, .44, 4444, 4.444).

.cat4! = .24

4[sup]sq[/sup]cat4 = 164

4cat4! = 424



We've been using .4[sup]sq[/sup] = .16, but in that case (.4)[sup]sq[/sup] = (.)(4[sup]sq[/sup])[/QUOTE]

Interesting suggestion cheesehead.

If that's possible, 164-1=163
164/.8=205

To put it another (and more precise) way, what is the precedence of the concatenation operation? I think we mostly assume that it has the highest precedence, so that 4.4[sup]sq[/sup] = (4.4)[sup]sq[/sup] = 19.36, not 4.4[sup]sq[/sup] = (4.)(4[sup]sq[/sup]) = 4.16 and that 44[sup]sq[/sup] = (44)[sup]sq[/sup] = 1936, not 44[sup]sq[/sup] = (4)(4[sup]sq[/sup]) = 416.

[quote=cheesehead]We've been using .4sq = .16, but in that case (.4)sq = (.)(4sq)[/quote]But for .4[sup]sqsq[/sup], ((.4)[sup]sq[/sup])[sup]sq[/sup] = .0256, while (.)((4[sup]sq[/sup])[sup]sq[/sup]) = .256

243 is out!
 

[QUOTE=cheesehead]To put it another (and more precise) way, what is the precedence of the concatenation operation? I think we mostly assume that it has the highest precedence, so that 4.4[sup]sq[/sup] = (4.4)[sup]sq[/sup] = 19.36, not 4.4[sup]sq[/sup] = (4.)(4[sup]sq[/sup]) = 4.16 and that 44[sup]sq[/sup] = (44)[sup]sq[/sup] = 1936, not 44[sup]sq[/sup] = (4)(4[sup]sq[/sup]) = 416.[/QUOTE]

Although the suggestion is pretty good, I wouldn't like to make things too easy...

243=(576/256)[sup]sq[/sup]*(4!+4!)

[i]13[/i] left!

Ah, so _that_'s how to get 3[sup]5[/sup].



[QUOTE=fetofs]Although the suggestion is pretty good, I wouldn't like to make things too easy...[/quote]Apparently not -- you've left 189 on the list so someone else can work on it. :-)

To complete through 300, these still need a solution
 

163
177
183
187
205
213
218
223
237
283
298

Hmmm... wasn't my list correct?

A few more solutions in 3/4, 2/4 and 1/4 time:

281 = 4[sup]sq[sup]sq[/sup][/sup]+4/.4[sup]sq[/sup]

304 = 4[sup]sq[sup]sq[/sup][/sup]+4!+4!

316 = 4[sup]sq[sup]sq[/sup][/sup]+4!/.4

640 = 4[sup]sq[sup]sq[/sup][/sup]/.4

784 = (4!+4)[sup]sq[/sup]

65536 = 4[sup]sq[sup]sq[sup]sq[/sup][/sup][/sup]

Yes, but you cannot edit it. As I see it, continuous repetition of "the (current) list" doesn't really add anything to the discussion, but rather detracts from it in a significant manner by obscuring the more meaningful part of the discussion. That is why I have been maintaining the list in a message and periodically re-posting it so that it recains near the end of the thread.
When I do so, I've deleted the previous version of the message.

My preference would have been to keep the "scoreboard" at the very top of the thread and update it in place. The next time one of these kinds of "puzzles" (amusements that lead to a large number of responses which in some way "refine" the answer) is submitted, I request that the person posting follow the procedure that I outlined earllier in this thread. That way we can have the information presented in a usable manner and avoid cluttering the content of the thread with "status" information.

[QUOTE=Wacky]That is why I have been maintaining the list in a message and periodically re-posting it so that it recains near the end of the thread.[/quote]Okay.

[quote]When I do so, I've deleted the previous version of the message.[/quote]Oh, I hadn't noticed. Makes sense, as does your recommendation. :-)

(edited for a couple mistakes - hope to replace soon)

In 2/4 time:

25 = 4/((.4)^2)

(of course, this means that there are lots of other ones that now are in 3/4 time, like 21=25-4, 29=25+4)

Oops. Just noticed Xilman's post tucked away there.

[QUOTE=rogue]I have an idea for a means to solve some of the remaining values (focusing more on the odd ones).

4 and 24, each require one 4
3, 5, 7, 11, 15, and 17, each require three 4s

Someone could write a program that could do repeated squarings of these numbers until the difference between two of them is 93. It shouldn't be hard to write, but it is possible that the number of squarings would exceed available memory on ones computer.[/QUOTE]

93=17^2 - 14^2.

I'm trying to use 93 = 47[sup]2[/sup]-46[sup]2[/sup], by first expanding the 4-4s representations of 47[sup]2[/sup] and 46[sup]2[/sup], then simplifying by collecting common terms. The expansion out to 39 fours worked just fine. So far I've simplified that down to only 32 fours:

93 = 4!*4!*4 - 4!*4 - 4!*4!/.4[sup]sq[/sup] + 4!*4/.4[sup]sq[/sup] + 4!*4/.4[sup]sq[/sup] - 4[sup]sq[/sup]/.4[sup]sq[/sup] + 4!*4/.4 + 4!*4/.4 - 4*4/.4 - 4*4/.4 - 4*4 + 4/4

I suspect that " + 4/4" term isn't going away, so I just have to reduce the other 30 fours down to a two-4s representation of 92.

[QUOTE=cheesehead]93 = 4!*4!*4 - 4!*4 - 4!*4!/.4[sup]sq[/sup] + 4!*4/.4[sup]sq[/sup] + 4!*4/.4[sup]sq[/sup] - 4[sup]sq[/sup]/.4[sup]sq[/sup] + 4!*4/.4 + 4!*4/.4 - 4*4/.4 - 4*4/.4 - 4*4 + 4/4[/QUOTE]Well, soon after I posted that I realized I could take out six 4s by combining the three terms just before "+ 4/4" to make another "- 4!*4" term:

4!*4!*4 - 4!*4 - 4!*4 - 4!*4!/.4[sup]sq[/sup] + 4!*4/.4[sup]sq[/sup] + 4!*4/.4[sup]sq[/sup] - 4[sup]sq[/sup]/.4[sup]sq[/sup] + 4!*4/.4 + 4!*4/.4 + 4/4

... and now we can collect all the terms involving 4! like so:

4!*(4!*4 - 4 - 4 - 4!/.4[sup]sq[/sup] + 4/.4[sup]sq[/sup] + 4/.4[sup]sq[/sup] + 4/.4 + 4/.4) - 4[sup]sq[/sup]/.4[sup]sq[/sup] + 4/4

= 4!*(4+4) - 4[sup]sq[/sup]/.4[sup]sq[/sup] + 4/4

but that may not have actually helped. :-}

93 = a^2 - b^2
 

I think that we can show that this is an impossible approach.

If 93 = a[sup]2[/sup] - b[sup]2[/sup]
then (a+b) | 93 and (a-b) | 93 .
Since 93 = 3 * 31, this leads to only two possible solutions,
namely 93 = 17[sup]2[/sup] - 14[sup]2[/sup]
and 93 = 47[sup]2[/sup] - 46[sup]2[/sup]
In both cases, the odd number cannot be generated by less than three fours. (No odd number can be generated in less than two).
But the even number cannot be generated by a single four.
Therefore, there is no solution of this form using at most four fours.

By similar logic, I beleive that we will also eliminate all other possible forms that conform to the set of "rules" that Ken chose for this thread.

[QUOTE=Wacky]In both cases, the odd number cannot be generated by less than three fours. (No odd number can be generated in less than two).

But the even number cannot be generated by a single four.
Therefore, there is no solution of this form using at most four fours.[/quote]Counterexample:

3 = 4-4/4

12 = 4[sup]sq[/sup]-4

15 = 4[sup]sq[/sup]-4/4

If a term in the fours representation of the odd number can be combined with a term in the fours representation of the even number to either reduce the number of fours necessary in the sum/difference or even cancel each other, then the representation of the sum/difference [u]can[/u] have fewer fours than the total number of fours in the two separate representations.

[quote]By similar logic, I beleive that we will also eliminate all other possible forms that conform to the set of "rules" that Ken chose for this thread.[/QUOTE]But we'll have to take the possibilities of simplification and cancellation into account.

Oh, wait -- you meant that in the [i]particular cases[/i] of 17[sup]2[/sup] and 14[sup]2[/sup] or 47[sup]2[/sup] and 46[sup]2[/sup], the odd numbers require at least three and the even numbers require at least two. So 3 + 12 = 15 is [i]not[/i] a valid counterexample.

I still don't see why it is not possible that cancellation or simplification could result in a four fours representation of 93 under the current set of rules.