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alexhak
2010-12-02 05:21:16

8/(3-(8/3))

krassi_holmz
2006-06-04 19:20:35

Here's the code (Mathematica, rewritten, but really messy and hard-to-understand):

#### Code:

```K[n1_, n2_] := Union[{n1 + n2, n1 - n2, n1*n2, n1/n2}];
KK[list_, num_] := Union[Flatten[Table[K[list[[i]], num], {i,
1, Length[list]}]]];
KKK[list1_,
list2_] :=
Union[Flatten[Table[KK[list1, list2[[i]]], {i, 1, Length[list2]}]]];
d[a_, b_, c_, d_, f_] := {
(*abcdfff*)
f[{a}, f[{b}, f[{c}, {d}]]],
(*abcfdff*)
f[{a}, f[f[{b}, {c}], {d}]],
(*abcffdf*)
f[f[{a}, f[{b}, {c}]], {d}],
(*abfcdff*)
f[f[{a}, {b}], f[{c}, {d}]],
(*abfcfdf*)
f[f[f[{a}, {b}], {c}], {d}]
}
d[l_, f_] := d[l[[1]], l[[2]], l[[3]], l[[4]], f];
dd[l_, f_] := dd[l[[1]], l[[2]], l[[3]], l[[4]], f];
dd[a_, b_, c_, d_, f_] :=
(
Print["abcdfff:", f[{a}, f[{b}, f[{c}, {d}]]]];
Print["abcfdff:", f[{a}, f[f[{b}, {c}], {d}]]];
Print["abcffdf:", f[f[{a}, f[{b}, {c}]], {d}]];
Print["abfcdff:", f[f[{a}, {b}], f[{c}, {d}]]];
Print["abfcfdf:", f[f[f[{a}, {b}], {c}], {d}]];
)
p = Permutations[{3, 3, 8, 8}];
res = Table[Union[Flatten[d[p[[i]], KKK]]], {i, 1, Length[p]}];
Union[Flatten[res]]```

Could explain and rewrite it later.

krassi_holmz
2006-06-04 19:10:31

This program was personal challenge.
Here's list of all numbers, which can be expressed using 8,8,3,3:

krassi_holmz
2006-06-04 19:08:17

My program is ready. And guess what-there aren't another solutions except:
8/(3-(8/3))=24!!!

krassi_holmz
2006-06-04 18:34:43

Making progress...

krassi_holmz
2006-06-04 17:27:08

Interesting... Can't you make some program, which gives all possible solutions?

justlookingforthemoment
2006-06-04 12:35:15

#### Zach wrote:

You realise that answer was said all the way up there, right?

And plus, you realise that this post is one year old? Almost exactly.

Oh well, well done for working it out. You missed a closing parenthesis in the second line though.

jebediah
2006-06-04 09:44:13

8/(3-(8/3) =
8/(3-(2 2/3) =
8/(1/3) = 24

Zach
2005-06-03 03:25:26

You realise that answer was said all the way up there, right?

JD
2005-06-02 22:12:18

Well, the answer is not easy, but it exists:

8/3=2,6666666666666666666... (=2,66periodic) (=8/3)

3-2,66p=0,33p (in other way, 9/3-8/3=1/3)

8/0,33p=24 (in other way, 8/1/3=8*3=24)

Summarizing...

8/(3-(8/3))=24

justlookingforthemoment
2005-05-31 18:10:10

Okie dokie, fixed it... I'm using firefox.

MathsIsFun
2005-05-31 18:06:02

Oh... I wonder why not. It is underlined in Internet Explorer and Firefox on my PC.

Do you have anything unusual about your setup? Operating System / Browser or something?

justlookingforthemoment
2005-05-31 17:14:27

Ohh! I see. Is there are way for you to make all links colourful or underlined or something in your forum? You can never tell if it is linked or not... or is that just me?

MathsIsFun
2005-05-30 21:08:23

"Click and ye shall find"

Roraborealis
2005-05-30 20:30:58

That page.