This continues Damiano's puzzle "Four dice puzzle: 2,2,4,5"
Damiano keeps throwing his four dice. After a lot of throwing and thinking and working, he has determined for every throw $a,b,c,d$ of his four dice the smallest positive integer $N(a,b,c,d)$ that cannot be generated from this throw according to the following rules:
- One may use the four numbers $a,b,c,d$ in any order, and it is fine if not all of them are used.
- Concatenation of digits is NOT allowed.
- The only allowed operations are additions, subtraction, multiplication, and division ($+,-,*,/$).
- One may use any number of brackets.
Question: Which throw $a,b,c,d$ of dice yields the largest number $N(a,b,c,d)$ ?
Answer
I have an answer of
2, 4, 5, 6
Which yields an N of
45
Barring a bug in my test program, this should be correct. I tried all possible values.
Here are all the values for my answer:
1 = 5 - 4
2 = 2
3 = 5 - 2
4 = 4
5 = 5
6 = 6
7 = 5 + 2
8 = 6 + 2
9 = 5 + 4
10 = 6 + 4
11 = 6 + 5
12 = 6 * 2
13 = 6 + 5 + 2
14 = (5 * 4) - 6
15 = 6 + 5 + 4
16 = (6 * 2) + 4
17 = (6 * 2) + 5
18 = (5 + 4) * 2
19 = (6 * 4) - 5
20 = 5 * 4
21 = ((6 * 4) + 2) - 5
22 = (5 * 4) + 2
23 = (5 * 4) + (6 / 2)
24 = 6 * 4
25 = ((6 + 4) * 2) + 5
26 = (6 * 4) + 2
27 = ((6 * 4) + 5) - 2
28 = (5 + 2) * 4
29 = (6 * 4) + 5
30 = 6 * 5
31 = (6 * 4) + 5 + 2
32 = (6 * 5) + 2
33 = ((2 / 4) + 5) * 6
34 = (6 * 5) + 4
35 = ((6 / 2) + 4) * 5
36 = (4 + 2) * 6
37 = ((6 + 2) * 4) + 5
38 = ((5 + 2) * 6) - 4
39 = ((5 / 2) + 4) * 6
40 = (6 + 2) * 5
41 = ((4 + 2) * 6) + 5
42 = (5 + 2) * 6
43 = (6 * 4 * 2) - 5
44 = (6 + 5) * 4
45 = ???
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