View the solution at http://mathforum.org/ruth/four4s.puzzle.html
Now, I ask to do the same thing but with 3s', that is, could you write equations (using four or five 3s') that have the numbers from 0 to 100 as the answer?
I started with some numbers, I ask the readers to complete the following list:
0 = 33 - 33
1 = 33/33
2 = 3/3 + 3/3
3 = 3!/3 + 3/3
4 = 3!/3 + 3!/3
5 = 3 + 3/3 + 3/3
6 = 3*(3/3 + 3/3)
7 = 3 + 3!/3 + 3!/3
8 = 3*3 - 3/3
9 = 3*3 + (3 - 3)
10 = 3*3 + 3/3
11 = 3*3 + 3!/3
12 = 3*3 + 3 * 3/3
13 = (-3/3 + 3^3)*(3/3!) or 3*3 + 3/3 + 3
14 = 3*3 + 3!/3 + 3
15 =
16 =
17 = 3*3 + (3*3 - 3/3)
18 = 3^3 - 3*3
19 = 3^3 - (3*3 - 3/3)
20 =
21 =
22 =
23 =
24 = 3^3 - 3 * 3/3
25 = (3 + 3/3)! + 3/3
26 = 3^3 - 3/3
27 = 3^3 + (3 - 3)
28 = 3^3 + 3/3
29 =
30 = 3^3 + 3 * 3/3
31 = 3^3 + 3 + 3/3
37 = 33 + 3 + 3/3
38 =
39 =
40 =
41 =
42 =
43 =
44 =
45 =
46 =
47 =
48 =
49 =
50 =
51 =
52 =
53 =
54 = 3^3 + 3^3
55 =
56 =
57 =
58 =
59 =
60 =
61 =
62 =
63 =
64 =
65 =
66 =
67 =
68 =
69 =
70 =
71 =
72 = 3*3 *(3*3 - 3/3)
73 =
74 =
75 =
76 =
77 =
78 =
79 =
80 =
81 = 3 * 3 * 3 * 3
82 =
83 =
84 =
85 =
86 =
87 =
88 =
89 =
90 =
91 =
92 =
93 =
94 =
95 =
96 =
97 =
98 =
99 =
100 = 3 * 33 + 3/3
No comments:
Post a Comment