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24= 3x8
24=8/3
There must be some link...................
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
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(8*3)-3^0+8^0=24
I know that 3^0 is not the same as 3, but it was said that we need to use 8,8,3,3 - and as you can see every of these numbers are used. I though of it logically, that's why i did it in this way. ????
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I've got the answer...at least I think so....
((8 x 3!)/3)+8
= ((8 x 3 x 2 x 1)/3)+8
= (48/3)+8
= (16)+8
= 24
yay!!
Its about half past midnight, but I just thought of a genius answer (that doesn't have factorials) and I couldn't wait to share!
8/(3-(8/3))=
8/(1/3)=
24
I'm so happy!
So impressed with both solutions ... !
And I thought we were just being teased.
I might add this one to our "official" puzzle list
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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Gasp! The horror.
Boy let me tell you what:
I bet you didn't know it, but I'm a fiddle player too.
And if you'd care to take a dare, I'll make a bet with you.
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What's the official puzzle list?
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Like, this page.
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
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Huh?
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That page.
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
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"Click and ye shall find"
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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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?
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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?
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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Okie dokie, fixed it... I'm using firefox.
I just ticked underline links!
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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
You realise that answer was said all the way up there, right?
Boy let me tell you what:
I bet you didn't know it, but I'm a fiddle player too.
And if you'd care to take a dare, I'll make a bet with you.
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8/(3-(8/3) =
8/(3-(2 2/3) =
8/(1/3) = 24
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.
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Interesting... Can't you make some program, which gives all possible solutions?
IPBLE: Increasing Performance By Lowering Expectations.
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Making progress...
IPBLE: Increasing Performance By Lowering Expectations.
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My program is ready. And guess what-there aren't another solutions except:
8/(3-(8/3))=24!!!
IPBLE: Increasing Performance By Lowering Expectations.
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This program was personal challenge.
Here's list of all numbers, which can be expressed using 8,8,3,3:
Last edited by krassi_holmz (2006-06-03 21:17:46)
IPBLE: Increasing Performance By Lowering Expectations.
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Here's the code (Mathematica, rewritten, but really messy and hard-to-understand):
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.
IPBLE: Increasing Performance By Lowering Expectations.
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8/(3-(8/3))