Tuesday, 29 September 2015

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. 





Thursday, 3 September 2015

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