9
$\begingroup$

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 $N(a,b,c,d)$ ?

$\endgroup$
7
$\begingroup$

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 = ???

$\endgroup$
  • $\begingroup$ 45 = (6*6)+5+4 or 45 = (6+3) * 5 or am I missing something? $\endgroup$ – VenomFangs Oct 12 '15 at 19:16
  • 1
    $\begingroup$ @VenomFangs You are missing something. Neither of your 2 options can be done with the 4 numbers I used to get 1-44. For example, 6,6,5,4 can get you 45, but it won't get you 33, so it isn't the best option. $\endgroup$ – Joel Rondeau Oct 12 '15 at 19:20
  • $\begingroup$ Gotcha, the question was more in depth than I was giving it credit for. Thanks! $\endgroup$ – VenomFangs Oct 12 '15 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.