Classic problem: Four 4's - Page 3

Ken_g6 · 2005-09-29, 01:44

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.

S80780 · 2005-09-29, 01:52

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

grandpascorpion · 2005-09-29, 02:28

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

rogue · 2005-09-29, 02:36

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

grandpascorpion · 2005-09-29, 02:59

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.

S80780 · 2005-09-29, 04:14

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

S80780 · 2005-09-29, 05:52

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

S80780 · 2005-09-29, 10:36

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

grandpascorpion · 2005-09-29, 12:09

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

Wacky · 2005-09-29, 12:28

Where is "75" ?

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

grandpascorpion · 2005-09-29, 13:50

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

2005-09-29, 02:28	#25
grandpascorpion Jan 2005 Transdniestr 503 Posts	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 44sq. 62 = (4sqsq - 4 -4)/4 89 = ((4!/4)sq-.4)/.4 90 = (4!/4)sq/.4 91 = ((4!/4)sq+.4)/.4 Last fiddled with by grandpascorpion on 2005-09-29 at 02:32*

2005-09-29, 10:36	#30
S80780 Jan 2003 far from M40 5³ Posts	Glancing over the record, I saw that 154 has a 3 4's representation, so: 158 = 4!/(.4)^2+4+4 Last fiddled with by Wacky on 2005-10-03 at 22:13

2005-09-29, 12:28	#32
Wacky Jun 2003 The Texas Hill Country 3²·11² Posts	Where is "75" ? From above: 77 = (4 - 4/4)sqsq - 4 94 = (4/.4)sq - 4!/4 105 = (44/4)^2-4^2 Last fiddled with by Wacky on 2005-10-03 at 22:12

2005-09-29, 13:50	#33
grandpascorpion Jan 2005 Transdniestr 503 Posts	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 Last fiddled with by Wacky on 2005-09-29 at 14:32 Reason: Note: Now removed from pending list

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2005-09-29, 01:44	#23
Ken_g6 Jan 2005 Caught in a sieve 5·79 Posts	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.

2005-09-29, 01:52	#24
S80780 Jan 2003 far from M40 125₁₀ Posts	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

2005-09-29, 02:36	#26
rogue "Mark" Apr 2003 Between here and the 2×3²×347 Posts	Oops. It must have been that beer I had this evening. I don't know what I was thinking.

2005-09-29, 02:59	#27
grandpascorpion Jan 2005 Transdniestr 111110111₂ Posts	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.

2005-09-29, 04:14	#28
S80780 Jan 2003 far from M40 5³ Posts	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

2005-09-29, 05:52	#29
S80780 Jan 2003 far from M40 5³ Posts	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

2005-09-29, 12:09	#31
grandpascorpion Jan 2005 Transdniestr 503 Posts	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