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.
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.
Hi Dave,
ReplyDeleteLong time no speak!
178=(15 * (3*4)) - 2
Hi Lawrence,
DeleteThanks for this - it's always good to get a complete set!
No probs! I recall doing something similar at school - it was called the 4 4's. Similar to what you've done here, but you weren't allowed combine digits (i.e. using 44 wasn't an option) and had to use all 4 4s. As I recall we managed to do it all but my memory's not what it used to be...
Delete