Numbers 1,2,3,4,5 - 101 to 200

After recent posts, here's the second instalment of 'produce each number from 1 to 200 using 1,2,3,4,5 and basic maths functions'.  Here's the second half - from 101 to 200.

101 = (53*2) + (1+4)
102 = 51 * 2 * (4-3)
103 = 123 - (4 * 5)
104 = (25 * 4) + 1 + 3
105 = (4+3) * (2+1) * 5
106 = (54 * 2) - (3-1)
107 = (13 * 5) + 42
108 = (5^3 - 21) + 4
109 = 134 - 25
110 = (53 * 2) + (4*1)
111 = 135 - 24
112 = 132 - (4*5)
113 = (41 * 3) - (5*2)
114 = (54 + 3) * 2 * 1
115 = ((3 * 4 * 2) -1) *5
116 = (23 + 5 +1) *4
117 = (24 * 5) - (3*1)
118 = 125 - (3+4)
119 = 23 * 5 + (4 * 1)
120 = 1 * 2 * 3 * 4 * 5
121 = (2 * 3 * 4 * 5) + 1
122 = 154 -32
123 = 123 * (5 - 4)
124 = 134 - (2 * 5)
125 = (4^(5-1) / 2) - 3
126 = 125 + (4-3)
127 = 1 * 5^3 + (4-2)
128 = (35 * 4) - 12
129 = 153 - 24
130 = 5^3 + 4 + (2-1)
131 = (25 * 4) + 31
132 = (3! * 15) + 42
133 = (45 * 3) - (1 * 2)
134 = (2 * 5 * 13) + 4
135 = (45 * 3) * (2-1)
136 = ((45 + 1) * 3) -2
137 = (35 * 4) - (1 +2)
138 = (4^3 + 5) * 1 * 2
139 = 4! * 3! - (5 * (2-1))
140 = ((45 + 2) * 3) -1
141 = 135 + 4 + 2
142 = (52 * 3) - 14
143 = 135 +  (4 * 2)
144 = 142 + (5-3)
145 = 152 - (4 + 3)
146 = (4! * 3!) + 5 - (1+2)
147 = 153 - (4 + 2)
148 = (35 + 2) * 1 * 4
149 = (12 * 3 * 4) + 5
150 = 15 * (4 + (3*2))
151 = 145 + (3 * 2)
152 = (3^4 - 5) * 1 * 2
153 = (53 - 2) * (4-1) = 1! + 2! + 3! + 4! + 5!
154 = (5! + 34) * (2-1)
155 = 153 + (4-2)
156 = 1 * 4! * (5 + 3/2)
157 = (31 * 5) + (4/2)
158 = (2 * 3^4) - (5-1)
159 = 213 - 54
160 = 154 + (3 * 2)
161 = 153 + (4 * 2)
162 = 54 * 3 * (2-1)
163 = ((51 + 4) * 3) - 2
164 = 152 +  (4 * 3)
165 = (34 + (2-1)) * 5
166 = 5^3 + (42 -1)
167 = (34 * 5) + (1 * 2)
168 = 213 - 45
169 = 13^2 * (5-4)
170 = (4! * (5+2)) +  (3-1)
171 = (51 +2 + 4) * 3
172 = (45 -2) * (3 +1)
173 = (34 * 5) + 1 + 2
174 = 4! * 5 * (2-1)
175 = (3+4) * 25 * 1
176 = ((5! * 3/2) - 4) * 1
177 = 153 + 24
178 = 5! + (1 + 4!) * 2) + 3
179 = (3^2 * 4 * 5) -1
180 = 5 * 4 * 3^2 * 1
181 = (5 * 4 * 3^2) + 1
182 = 152 + 3! + 4!
183 = (54 * 3) + 21
184 = (5! * 3/2) + (4*1)
185 = 5! + 4^3 + (2-1)
186 = (5! + 4) * 3/2 * 1
187 = ((4^3) * (2+1)) - 5
188 = (32 * (5+1)) -4
189 = 31.5 * (4+2)
190 = (4! * (5+3)) - (1*2)
191 = (5 * 34) + 21
192 = 24 * (5+3) * 1
193 = 4! * (5*3) + 1
194 = (43 * 5) - 21
195 = (51 * 4) - 3^2
196 = (4 + 3)^2 * (5-1)
197 = ((52 -3) * 4) - 1
198 = (51 * 4) - (3*2)
199 = (51 * 4) - (3+2)
200 = (1 + 3 +4) * 25

I posted my results for 1-100 and my intention to post 101-200 on the freemathhelp forum (which I strongly recommend for any maths or logic problem or puzzle), along with a challenge to improve my answers with more elegant ones.  For example, using only basic maths function (no powers), and possibly answers which use the digits 1,2,3,4,5 in order.  One of the regular contributors there, Denis, took up my challenge and has produced solutions for 1-200 using the digits in correct order.  I had enough difficulty doing this with the digits in any order (occasionally stumbling on elegant solutions), but he's got the full set.  He's kindly shared his full solution with me, and permitted me to share it (after all, that's what problem solvers do), and that will be next time.

Numbers 1,2,3,4,5 from 1 to 200

Since pure and applied Maths are constantly expanding, it seems only fair that I should expand my tiny contribution to the field of puzzle solving.  My recent Maths posts have looked at the puzzle of making the numbers from 1 to 200 using only the four digits 1, 2, 3 and 4 and I've been very grateful for the support of the Maths community in solving this puzzle.  (1-100), (101-150), (151-200)

Extending this then, I set my mind to seeing if this can be done with the digits 1-5.  In some ways it's easier, in other ways it is more tricky. I've managed to solve for 1-200 (single-handedly, which suggests it's an easier task).  There is less need to employ factorials, and if I recall correctly, I've completed my solution without any decimals. 

Note
It is of course possible to take the four-number solution for (n-5) and attach a +5 at the end of the expression.  Or alternatively, any number between 121 and 320 can be achieved easily by adding 5! to the four-number solution. I have avoided these short-cuts (with one exception where I could not find an alternative - 178).

Here is part 1, going from 1 to 100:

1 = 1 * ((5+2)/(4+3))
2 = 1 + ((5+2)/(4+3))
3 = 3 * ((1+5)/(4+2))
4 = 4 * (1*5)/(2+3))
5 = 5 * ((4+1)/(2+3))
6 = (14*3) / (2+5)
7 = 35 / ((4+2) - 1)
8 = 32 / ((5+4) -1)
9 = (34+2) / (5-1)
10 = (1 + 24 + 5) / 3
11 = (15 - 4) * (3-2)
12 = 45/3 - (1+2)
13 = (4^2 - 5) + (3-1)
14 = (32/4) + 5 + 1
15 = (51-32) - 4
16 = (52 - 4) / (3 * 1)
17 = 25 - (4 + 3 + 1)
18 = 54/3 * (2-1)
19 = (54 + 1 + 2) / 3
20 = (42/3) + 5 + 1
21 = (54/3) + 1 + 2
22 = 32 - (1 + 5 + 4)
23 = (45 - 23) + 1
24 = 54/2 - (1*3)
25 = (52 + 4) -31
26 = (14 * (3/2)) + 5
27 = (53 -1) / (4-2)
28 = (54+2) / (3-1)
29 = (35-(4+2)) * 1
30 = (45 * 2)/(1 * 3)
31 = 5^2 + 4 + (2*1)
32 = (51+4) - 23
33 = 24 + 1+ 5 + 3
34 = 2^5 + 1 + (4-3)
35 = (152/4) -3
36 = (54+3) -21
37 = (5+1)^2 + (4-3)
38 = (132/4) + 5
39 = 1* 4^3 - 5^2
40 = 5! / ((12+3) / (1+4))
41 = (52+3) - 14
42 = (4*5*2) + (3-1)
43 = (125+4) /3
44 = (45*1) + (3-2)
45 = (54 - 3^2) * 1
46 = ((5*4) + (3*1)) * 2
47 = (35+14) -2
48 = (45+3) * (2-1)
49 = ((5+4+1)-3)^2
50 = (25*4)/(3-1)
51 = 34 + 15 + 2
52 = (4*5*1)+32
53 = (34*2) - 15
54 = 3^2 + 45 * 1
55 = (15*4) - (3+2)
56 = (2*4) * (5 + (3-1))
57 = (13*5) - (2*4)
58 = (25+4) * (3-1)
59 = (3*4*5) - (2-1)
60 = 1 + 24 + 35
61 = (1* 4^3) - (5-2)
62 = 13*4 + (2*5)
63 = (34 * 2) - (5*1)
64 = 3^4 - (15 + 2)
65 = (54 + 13) -2
66 = (35 * 2) - (4 *1)
67 = (24*3) - (5*1)
68 = (42+31) - 5
69 = 132 - 54
70 = (25 * 3) - (1*4)
71 = (43 * 2) -15
72 = (34*2) + (5-1)
73 = 3^4 - (5+2+1)
74 = 4^3 + (1 + 5 + 2)
75 = (35*2) + 1 + 4
76 = ((35 + 4) -1) * 2
77 = (3+4) * ((5*2) +1)
78 = (25 * 3) + (4-1)
79 = ((5^2) * 3))+(4*1)
80 = (2*41) - (5-3)
81 = (5+4) * 3^2 * 1
82 = 134 - 52
83 = (4^2 * 5) + (3*1)
84 = (35 * 2) + 14
85 = (5! - 32) - (4-1)
86 = (21 * 4) + (5-3)
87 = 54 + 31 + 2
88 = (45 * 2) - (3-1)
89 = (13 * 5) + 24
90 = (41 + 52) - 3
91 = 13 * ((5+4)-2)
92 = (45 * 2) + (3-1)
93 = (14 * 5) + 23
94 = (45 * 2) + 1 + 3
95 = (51 * 2) - (3+4)
96 = (45 + 3) * 1 * 2
97 = (25 * 4) - (3 * 1)
98 = (41 * 3) - 25
99 = (21 * 4) + (3 * 5)
100 = ((53 - 4) + 1) * 2

Next time: 101 - 200. 

Numbers 1,2,3,4 from 151 to 200

After my previous posts on using the digits 1,2,3,4 and mathematical operators (which, by the way, have become increasingly creative and powerful) to create the numbers 1-50, then 51-100 and 101 to 150, I'd like to present a team effort on the numbers from 151 to 200.  I didn't think it was possible.  In fact, I very much doubt that the authors of the maths textbook that posed the original idea thought it would be possible.  Nonetheless, here it is:  use the digits 1,2,3,4 and any mathematical operators you care to name to produce the totals from 151 to 200.

Credit to Denis from the freemathhelp forum for his work on this, providing the vast majority of the results and taking the problem way beyond its original scope.  Major credit also to Skipjack for providing the solutions marked with an asterisk.

To clarify, r(.1) = repeating decimal.

151: (4! + 3!) / .2 + 1

152: (1 + 4)! + 32

153: 34 / 2 / r(.1)

154: 2^4 / .1 - 3!

155: 31 * 2 / .4

156:4! * 3! + 12

157: 314 / 2

158: 3!! * .2 + 14

159: 3!! / 4 - 21

160: 32 * (1 + 4)

161: 3^4 * 2 - 1

162: 3^4 * 2 * 1

163: 3^4 * 2 + 1

164: (3! - 2) * 41

165: 4! + 3! + 21

*166: 4! * (3! + 1) - 2

167: 13^2 - SQRT(4)

168: 14 * 3! * 2

169: 13^(4 - 2)

170: 34 / .2 * 1

171: 13^2 + SQRT(4)

172 = .1^(-2) + 3 * 4!

173: 13^2 + 4

*174: 3! * (4! + 1 / .2)

175: 3!! / 4 - 1 / .2

*176: 4! / .1 - 2^3!

177: 3!! / 4 - 1 - 2

178: 3!! / 4 - 1 * 2

179: 3!! / 4 + 1 - 2

180: 3!! / 4 * (2 - 1)

181: 3!! / 4 - 1 + 2

182: 3!! / 4 + 1 * 2

183: 3!! / 4 + 1 + 2

184: 3!! * .2 + 4 / .1

185: 3!! * .2 + 41

186: (2 + 4) * 31

*187: (4! - 3) / r(.1) - 2

188 = (.1^(-2) - 3!) * SQRT(4)

189: 213 - 4!

190: 14^2 - 3!

191: 4! * 2^3 - 1

192: 3!! / 4 + 12

193: 14^2 - 3

194: 4 / .1 / .2 - 3!

195 = (4! - 2) / r(.1) - 3

*196: 4 * (3! + 1)^2

197: 4 / .1 / .2 - 3

198 = (3! * 4 - 2) / r(.1)

199: 14^2 + 3

200: (2 + 3) * 4 / .1