Friday, 23 October 2015

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.











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