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#1 2008-06-12 09:03:59

andrew21
Member
Registered: 2008-06-12
Posts: 2

probably the hardest puzzle ever

this is seriously the hardest maths/logic puzzle i have every seen and im no dunce when it comes to maths straight As but this one i was working on for 2 hours and no luck at all but am completely certian its possible so id thought id try and search the web and see if i can find a solution there.

the rules are that you need to combine any of the letters below so that each number appears an odd number of times exept for 0 which has to appear an even number of times.

may be a bit hard to understand what i mean but hopefully most people will understand.

also i dont suggest you attempt this if unless you are insanely intelligent or have alot of time on your hand or preferably both as i dont wish to waste peoples time.

a)0,1,2,3,4,5
b)0,1,2,4,7,8
c)0,1,2,3,4,5,9,10,11
d)0,2,3,5,6,12
e)0,1,2,4,7,8,9,10,11,13
f)0,2,3,4,6,9,11,12
g)3,5,6,11,12,14
h)1,4,7,8,15,16
i)1,4,7,8,10,13,15,16,23
j)2,4,5,9,10,11,17,18,19
k)2,4,8,9,10,17,19,20
l)2,5,6,9,11,14,17,18,21
m)3,5,6,12
n)4,8,10,13,16,19,20,22,23
o)5,6,11,14,18,21
p)7,8,15,16
q)7,8,13,15,16,23
r)9,10,11,17,,18,19,24,25,26
s)9,11,14,17,18,21,25,26,27
t)9,10,13,17,19,20,24,26,28
u)10,13,19,20,22,23,24,28,29
v)11,14,18,21,25,27
w)13,20,22,23,28,29
x)8,13,16,20,22,23
y)17,19,20,24,26,28,30,31,32
z)17,18,21,25,26,27,30,33,34
A)17,18,19,24,25,26,30,33                      yes it keeps going
B)18,21,25,27,33,34
C)19,20,22,24,28,29,31,32,35
D)20,22,28,29,31,35
E)24,25,26,30,32,33
F)24,28,29,31,32,35
G)24,26,28,29,30,31,32
H)25,26,27,28,30,33
I)25,27,33,34
J)28,29,31,35




like i said insanely hard but althought there are alot of letters to choose from 36 in total it may (enthasis on the may) not be as difficult as it looks as only letter need only be used once twice and it would effectively cancle itself out and just to help a few people heres i few things i noticed which may help

tips
most letters will never be used so dont be intimidated by the number its just a large selection to choose from thats all
start with high or low numbers then work your way up or down
theres alot of combinations for the higher numbers in the letters that can quickly get rid of a large portion of the numbers quickly q,z C seem to get rid of alot using just 3

if any one can complete this i will be massively surprised but very pleased big_smile i will post the answer to this as soon as i get it (if i ever get it)

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#2 2008-06-12 10:48:38

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: probably the hardest puzzle ever

A fairly powerful (I think) trick is to create new letters by combining existing ones.
As Andrew says, if a number appears twice in a letter then it's the same as if it's not there, and so if two fairly similar letters are chosen to combine then you could end up with a letter that contains less numbers than either of its components.

A letter with a low amount of numbers is useful, because it allows for more flexibility. In particular, if you find a way to make a letter containing exactly one number, then you can ignore all instances of that number in the puzzle (because you will always be able to add or remove that number as needed).

Finding a way to express all numbers individually would be sufficient (but not necessarily necessary) to solve the puzzle.

Just scanning this quickly for an example:

p+q = 13,23
h+i = 10,13,23

∴ h+i+p+q = 10.
So now the puzzle can be continued as if 10 isn't there.

Mostly I can only find ways to make pairs though. Still useful, but not nearly as much as singles.


Why did the vector cross the road?
It wanted to be normal.

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#3 2008-06-13 00:29:13

andrew21
Member
Registered: 2008-06-12
Posts: 2

Re: probably the hardest puzzle ever

do any of these combos you get of just 2 contain  0 3 5 12 or 0 6 11 12 as i can do all apart from them if u can tell me i will be able to post a solution to the problem


edit:managed to get it to 9 and 2 if any one can get those 2

Last edited by andrew21 (2008-06-13 00:34:52)

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#4 2008-06-13 04:39:43

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: probably the hardest puzzle ever

Finally done!

I managed to express each number individually, and so by adding them you can make any combination of numbers possible.

 0 = fjmuwACDE
 1 = inqxCDFJ
 2 = dfjuwACDE
 3 = cfhmnpxCDFJ
 4 = hinpqxCDFJ
 5 = acdfnxCDFJ
 6 = acdfgmnoxBCDFIJ
 7 = efghjknprtvxBCDEHI
 8 = bdefgjkmnrtvxBCDEHI
 9 = abcdefjlopqstvwyADEFG
10 = hipq
11 = bdefhijlostvwyADEFG
12 = cfghmnopxBCDFIJ
13 = abce
14 = bdefghijlmostvwyADEFG
15 = efghjknpqrtvBCEHIJ
16 = bdefgjkmnpqrtvBCEHIJ
17 = abcepqstuvyCFG
18 = abcehistvwyACEGJ
19 = CDFJ
20 = bdefgjlmortvBCDEHI
21 = abcehistvwyABCEGIJ
22 = bdefgjlmortvBCEHIJ
23 = abcdpq
24 = hipquwCDFJ
25 = acdfjlorsuvwCDFJ
26 = dfjlosuvwACDE
27 = acdfgjlmorsuwBCDFIJ
28 = acdfgjlmorsuwBCDEHI
29 = acdfgjlmopqrsuBCEHIJ
30 = abcdefjlopqtwyzABDEFG
31 = abcestuvwyzBCD
32 = hipquwCD
33 = bdefjlopqrtwyzBDEFG
34 = bdefgjlmopqrtvwyzDEFG
35 = abcepqstuvyzBC

This can be used to find the answer, and unless I've made a mistake somewhere, the winning combination should be abcegilopqstuvyzFG.


Why did the vector cross the road?
It wanted to be normal.

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