thePhantom

150438. Thu Feb 22, 2007 1:53 pm 


here is a maths game that's Qi ...
Use the digit 4, up to four times to make 1,2,3 … etc
Using: + , ÷ , × ,  , √ , ( ) , ( )^2 and ! (as many times as necessary)
lets see how far we can get!
i'll go first ... 1= 4 ÷ 4
[There now follows a telescoped version of the numbers up to 183, which was as far as this thread got. QIM.)
2 = (4+4) /4 (djg)
3 = 4  (4/4) (Zaphod)
4 = 4 (ameena)
5 = 4 + (4/4) (samivel)
6 = 4 + √4 (Zaphod)
7 = 4 + 4 4/4
8 = 4 + 4
9 = 4 + 4 + 4/4 (djg)
10 = 4 + 4 + √4
11 = 44 ÷ 4 (suze)
12 = 4 + 4 + 4 (djg)
13 = 4!  44/4 (suze)
14 = 4!  (4+4+√4) (ali)
15 = (4 x 4)  (4/4)
16 = 4 x 4
17 = (4 x 4) + (4/4) (djg)
18 = 4!(4+√4)
19 = 4!(4+4/4)
20 = 4x4+4 (ali)
21 = 4! + (4/4)  4 (suze)
22 = 4!  ((4+4)/4) (ali)
23 = 4!  4/4
24 = 4!
25 = 4! + 4/4 (ali)
26 = 4! + ((4+4)/4) (djg)
27 = 4! + 4  4/4
28 = 4! + 4
29 = 4! + 4 + 4/4 (ali)
30 = 4! + 4 + √4 (djg)
31 = 4! + ((4 + 4!)/4)
32 = 4 x 4 x √4
33 = 4! + 4 + (√4/.4) (suze)
34 = (√4^4)x√4+√4 (ali)
35 = 4! + (44/4)
36 = 4! + 4 + 4 + 4
37 = 4! + ((4! + √4) ÷ √4) (suze)
38 = 4! + 4^2  √4
39 = 4! + 4^2  4/4
40 = 4! + 4^2
41 = 4! + 4^2 + 4/4
42 = 4! + 4^2 + √4 (djg)
43 = 44  4/4
44 = 44
45 = 44+ 4/4
46 = 44 + √4
47 = 4! + 4!  4/4
48 = 4! + 4!
49 = 4! + 4! + 4/4
50 = 44+ (4!/4) (suze)
Not being expert in the higher branches of math, I don't know what a gamma function is (husband and s/d were quite prepared to tell me, but I didn't insist on it). But all the same, the fact that Γ(4) = 6 is apparently useful for some of the larger numbers. suze.
The gamma function is an extension of the factorial function to real and complex arguments. For our purposes though Γ(n)=(n1)! ali.
51 = √4x4!+ Γ(4)/√4
52 = √4x4!+4
53 = √4x4!+√4/.4
54 = 44+4/.4
55 = 44/(√4x.4)
56 = (4+4/.4)x4
57 = 4!/.4  Γ(4)/√4 (ali)
58 = (4!/.4)  √4
59 = (4!/.4)  4/4
60 = (4!/.4) (rat)
61 = (4!/.4) + 4/4
62 = (4!/.4) + √4
63 = ((4^4)4) ÷ 4
64 = √4^Γ(4) (suze)
65 = 4^3 + 4/4
66 = 4^3 + √4
67 = 4^3 + 4 4/4
68 = 4^3 + 4
69 = 4^3 + 4 + 4/4 (djg)
70 = 4^3 + √ 4 + 4
71 = 4^3 + Γ(4) + 4/4
72 = 4^3 + 4 + 4 (v/r)
73 = (4!/(((.4)^2)x√ 4))√ 4
74 = √ 4^Γ(4)+4/.4
75 = 4!/(((.4)^√ 4)x√ 4)
76 = 4!/.4 + 4^√4
77 = (4!/((.4^2)x√4) + √4
78 = (4!xΓ(4))/√4 + Γ(4)
79 = (4!/((.4^2) x √4) + 4
80 = √4^Γ(4) + 4^√4
81 = (Γ(4) + Γ(4)/√4)^√4 = (Γ(4)/√4)^4 (ali)
82 = 4^2 + 4^3 + √4 (v/r)
83 = (Γ(4)/√4)^4) + √4
84 = 44x√4  4
85 = Γ(Γ(4))  ((4^2)√4)/.4
86 = 44x√4  √4
87 = Γ(4)/.4 + (Γ(4)^2)x√4 (ali)
88 = 44 x √4 (djg)
89 = (Γ(4)xΓ(4)  .4)/.4 (ali)
90 = (44 x √4) + √4 (djg)
91 = (Γ(4)xΓ(4) + .4)/.4
92 = 44x√4 +4
93 = 4x4!  Γ(4)/√4 (ali)
94 = 44x√4 + Γ(4)
95 = (Γ(4)xΓ(4)+√4)/.4
96 = 4x4! (ali)
97 = (4x4!)+(4÷4)
98 = (4x4!)+√4 (thePhantom)
99 = 4x4! + Γ(4)/√4
100 = 4x4! + 4
101 = 4x4! + √4/.4
102 = 4x4! + Γ(4) (ali)
103 = 4^4 x .4 + .Γ4
104 = .4(4^4 + 4) (Minivoltaire)
105 = Γ(Γ(4))  (4!/(.4*4))
106 = 4x4! +4/.4
107 = Γ(Γ(4))  ((4! + √4) / √4)) (ali)
108 = Γ(Γ(4))(4+4+4)
109 = Γ(Γ(4))  ((4!  √4) / √4))
110 = Γ(Γ(4))  (4*4) + Γ(4) (Minivoltaire)
111 = (44+.4)/.4 (ali)
112 = Γ(Γ(4))  4 * √4
113 = Γ(Γ(4))  4 * √4 + Γ(√4)
114 = Γ(Γ(4))  4 * √4 + √4
115 = Γ(Γ(4))  (4 + Γ(√4))
116 = Γ(Γ(4))  4
117 = Γ(Γ(4))  4 + Γ(√4)
118 = Γ(Γ(4))  4 + √4
119 = Γ(Γ(4))  Γ(√4)
120 = Γ(Γ(4)) (Minivoltaire)
121 = Γ(Γ(4)) + 4/4
122 = Γ(Γ(4)) + √4
123 = Γ(Γ(4)) + Γ(4)/√4
124 = Γ(Γ(4)) + 4
125 = Γ(Γ(4)) + √4/.4
126 = Γ(Γ(4)) + Γ(4) (ali)
127 = Γ(Γ(4)) + 4 * √4  Γ(√4)
128 = Γ(Γ(4)) + 4 * √4 (suze)
My limited math knowledge is beginning to fail me now, specially since my husband informs me that a double factorial is needed to do 129. 4!! is apparently 8, although I didn't really understand the explanation. suze.
129 = 4^4/√4 + Γ(√4)
130 = Γ(Γ(4)) + 4 * √4 + √4
131 = Γ(Γ(4)) + 44/4
132 = Γ(Γ(4)) + 4 + 4 + 4 (Minivoltaire)
133 = Γ(Γ(4)) + (4!+√4)/√4
134 = (4^4/√4)+Γ(4)
135 = Γ(Γ(4)) + Γ(4)/.4
136 = Γ(Γ(4)) + 4*4 (ali)
137 = Γ(Γ(4)) + 4*4 + Γ(√4)
138 = Γ(Γ(4)) + 4*4 + √4
139 = 4! * Γ4  4  Γ(√4)
140 = 4! * Γ4  4
141 = 4! * Γ4  4 + Γ(√4)
142 = 4! * Γ4  4 + √4
143 = 4! * Γ4  Γ(√4)
144 = 4! * Γ4
145 = 4! * Γ4 + Γ(√4)
146 = 4! * Γ4 + √4
147 = 4! * Γ4 + 4  Γ(√4)
148 = 4! * Γ4 + 4
149 = 4! * Γ4 + 4 + Γ(√4)
150 = 4! * Γ4 + 4 + √4
151 = 4! * Γ4 + Γ4 + Γ(√4) (Minivoltaire)
152 = 4! * Γ(4) + 4*√4
153 = 4! * Γ(4) + 4!! + Γ(√4)
154 = 4! * Γ(4) + 4/.4
155 = (√4^Γ(4)  √4)/.4
156 = 4! * Γ(4) + Γ(4) + Γ(4) (ali)
n!! = 1 iff n =1,0 or 1; n!! = n*(n2)!! iff n >= 2. ali.
This one is difficult and I'm going to have to introduce another function; the Hyperfactorial H(n), where H(n) = 1^1*2^2*...*n^n. ali.
157 = H(Γ(4)/√4) + Γ(4)!! + Γ(√4); to make it a bit clearer: H(Γ(4)/√4) = H(3) = 108; Γ(4)!! = 6!! =6*4*2 = 48; Γ(√4) = 1 (ali)
158 = 4(triangled)^√4 + Γ(4)!! + 4 (triangled) [4(triangled) = 4 + 3 + 2 + 1 = 10; henceforward T(4)]
159 = T(T(4) + 4!! + Γ(√4))  Γ(4) (Minivoltaire)
160 = T(4) x 4 x 4 (suze)
161 = (√4^Γ(4))/.4 + Γ(√4)
162 = (√4^Γ(4))/.4 + √4
163 = T(T(4) + 4!!  Γ(√4)) + T(4) (ali)
164 = (T(4) x 4 x 4) + 4 (suze)
165 = T(T(4)) + 44/.4
166 = (T(4) x 4 x 4) + Γ(4)
167 = Γ(Γ(4)) + T(T(4))  4!!
168 = (√4^Γ(4))/.4 + 4!!
169 = 4!*Γ(4) + T(4)/.4
170 = (T(Γ(4))  4)*T(4) (ali)
171 = T(T(4) + 4!!)
172 = T(T(4) + 4!!) + Γ(√4)
173 = T(T(4) + 4!!) + √4
174 = ((4 + Γ(√4))! + 4)sub[4!!] (Minivoltaire) [i.e. it's in base 8]
There was some discussion as to whether or not this was valid, especially since that calculation produces 174 (base 8), which is 124 (base 10). Eleanor (suze's stepdaughter) came up with an alternative, based on something she'd seen in a book.
This relies on what is called the floor function, meaning the largest integer n s.t. n is not greater than x. The usual way to denote this requires symbols not available here, so I've used square brackets. For instance [3.14159] = 3.
It's a bit long winded so I shall spell it out. We first need to know that 30 can be expressed with one 4; thus:
30 = [√√√√((4!)!)] i.e. the integer part of the sixteenth root of 24 factorial.
We then note that [√([√√√√((4!)!)])] i.e. the integer part of √30 is 5. Next, we note that ([√([√√√√((4!)!)])])! is 120. Then, [√(([√([√√√√((4!)!)])])!)] is 10, and so [√([√(([√([√√√√((4!)!)])])!)])] is 3.
Now,
174 = ([√([√√√√((4!)!)])])! + 4! + 4! + ([√([√(([√([√√√√((4!)!)])])!)])])! (Eleanor)
suze here again. While Eleanor was explaining that, I did a bit of Googling and found that this approach can be used to form every number up to at least 316. But is it getting too contrived?
174 = (T(Γ(4))  4)*T(4) + 4 (ali)
175 = ([√([√√√√((4!)!)])])! + [√√√√((4!)!)] + 4! + [√√4] (120 + 30 + 24 + 1)
176 = ([√([√√√√((4!)!)])])! + [√√√√((4!)!)] + 4! + √4
177 = ([√([√√√√((4!)!)])])! + [√√√√((4!)!)] + 4! + [√([√(([√([√√√√((4!)!)])])!)])]
178 = ([√([√√√√((4!)!)])])! + [√√√√((4!)!)] + 4! + 4
179 = ([√([√√√√((4!)!)])])! + [√√√√((4!)!)] + 4! + [√([√√√√((4!)!)])]
180 = 4! x 4!!  (√4 x Γ(4)) (suze)
181 = 4^4  ([√√√√((4!)!)]/.4)
182 = 44 x 4 + Γ(4)
183 = [(√√4}^(Γ(4)/.4)] + √4 (ali)
This was far as we got! 




bobwilson

430711. Tue Oct 28, 2008 9:18 pm 


hang on
the original post said Quote:  Use the digit 4, up to four times to make 1,2,3 … etc
Using: + , ÷ , × ,  , √ , ( ) , ( )^2 and ! (as many times as necessary) 
so where did this gamma function come in? 




suze

430902. Wed Oct 29, 2008 8:27 am 


Yea, the original rules were expanded a bit in order to make some of the numbers. As well as the gamma function, we resorted to double factorials, hyperfactorials, triangle functions, and floor functions, even though we balked a bit at bases.
If anyone can make any of the numbers which use those abstruse functions in a simpler way, please do! 




cherrycoke

436682. Sat Nov 08, 2008 1:10 am 


Well how about, for example rather than
175 = ([√([√√√√((4!)!)])])! + [√√√√((4!)!)] + 4! + [√√4] (120 + 30 + 24 + 1)
you take
10 = 4 + 4 + √4
17 = (4 x 4) + (4/4)
5 = 4 + (4/4)
so 175 = (((4 x 4) + (4/4)) * (4 + 4 + √4)) + (4 + (4/4))
or (17 * 10) + 5
Or am i missing a rule that says you can't do this? 




samivel

436685. Sat Nov 08, 2008 2:28 am 


You're only allowed to use the digit 4 four times. 




cherrycoke

436707. Sat Nov 08, 2008 4:38 am 


ah, must read more carefully 




BeccaD

437102. Sun Nov 09, 2008 6:02 am 


184 isn't too tricky: ((4! + (4)^2)) x 4) + 4!
if you use 4's as a power you could get one with 4^(44) but then you've only got a 4 left, so unless you're looking for 4 that's not very helpful. And even then you can just use 4. Or 4/4. Doh. and there's me trying to be clever. 




samivel

437177. Sun Nov 09, 2008 8:22 am 


You've used a 2 though, and that's not'lowed. 




suze

437188. Sun Nov 09, 2008 9:27 am 


Ordinarily it wouldn't be, but the rules for the problem as originally specified by thePhantom did say that squaredses were allowed. (As I was reminded when I called djg on the exact same thing back in the day!) 




ali

437201. Sun Nov 09, 2008 9:43 am 


samivel wrote:  You've used a 2 though, and that's not'lowed. 
I'm inclined to agree, but the fact is '^2' has been used many times before...
It can be done without, however as:
184 = Γ(Γ(4)) + (√4^Γ(4))
185 = Γ(Γ(4)) + (√4^Γ(4)) + Γ(√4) 




samivel

437619. Mon Nov 10, 2008 3:23 am 


Ah, it's a square, is it? Curse this nonsuperscripted computer! 




Sophie.A

628213. Wed Oct 21, 2009 9:41 am 


186 = Γ(Γ(4)) + (√4^Γ(4)) + √4
187 = {[Γ(4)]! + 4!} / 4 + Γ(√4)
188 = [4! × √4 − Γ(√4)] × 4 




ali

628222. Wed Oct 21, 2009 10:01 am 


189 = (Γ(4))!! * 4  T(√4)
190 = (Γ(4))!! * 4  √4
191 = (Γ(4))!! * 4  Γ(√4)
192 = (Γ(4))!! * 4
193 = (Γ(4))!! * 4 + Γ(√4)
194 = (Γ(4))!! * 4 + √4
195 = (Γ(4))!! * 4 + T(√4)
196 = (Γ(4))!! * 4 + 4 




ali

628263. Wed Oct 21, 2009 10:56 am 


Sophie.A wrote:  186 = Γ(Γ(4)) + (√4^Γ(4)) + √4
187 = {[Γ(4)]! + 4!} / 4 + Γ(√4)
188 = [4! × √4 − Γ(√4)] × 4 
Just to avoid potential confusion  we've been using square brackets above to represent the floor function, but here they are just used to make the parenthesis nesting clearer. 




Sophie.A

628291. Wed Oct 21, 2009 12:36 pm 


I see. Okay, I’ll avoid square brackets.
197 = φ(4!) × 4! + 4 + Γ(√4)
198 = φ(4!) × (4! + Γ(√4)) − √4
199 = φ(4!) × (4! + Γ(√4)) − Γ(√4)
200 = φ(4!) × (4! + Γ(√4))
201 = φ(4!) × (4! + Γ(√4)) + Γ(√4)
202 = φ(4!) × (4! + Γ(√4)) + √4
Where φ is the Euler totient function: for each positive integer n, φ(n) denotes the number of positive integers not greater than n that are coprime with n. This is a very commonly used function – any mathematics undergraduate student doing a course in number theory should be familiar with it. Last edited by Sophie.A on Thu Oct 22, 2009 7:25 am; edited 1 time in total




