
Content Count
806 
Joined

Last visited

Days Won
11
Content Type
Profiles
Forums
Calendar
Gallery
Blogs
Posts posted by wolfgang


As the numbers are from 1 to 6 that means it may have a relation to dice ( by using two dice):
11 means A
12 means B
21 means C
something like that...

It has something to do with names, but I have no idea.

The common number between 30 and 18 is 90.
So we bundle the sausages together and cut them into 3 equal parts, resulting 90 pieces.
90/18= 5 pieces for each person.
SpoilerIt looks simple

we can give him
Spoilerrightdownrightdown.....and so on
so that the number of orders= n+1

Is it possible to use a.....
Spoilerrope?

(BR / BL +UR) UL
Thus...
(8/4 +2) 3= 12
(2/1 + 6) 4= 32
(9/3 +4) 2= 14
and for the last one..
(6/2 +3) 5 = 30
 1

The Big Bang theory is the prevailing cosmological model for the universe from the erliest known periods through its subsequent largescale evolution.The model accounts for the fact that the universe expanded from a very high density and high temperature state Some estimates place this moment at approximately 13.8 billion years ago, which is thus considered the (age of the universe) After the initial expansion, the universe cooled sufficiently to allow the formation of subatomic particles, and later simple atoms. Giant clouds of these primordial elements later coalesced through gravity to form stars and galaxies.
My question was so: what if the whole universe was inside a colsed room? and the galaxies were just like dust particles hanging there , if these paricles are subjected to a continuous diverging gravity ,so they will expand.
That means , that there was NO start point and NO Big bang theory !
What do you think?

We should devide the 12 coins into three groups, 5,5,and 2
1 for the first five, we weigh 2 against 2, leaving one coin aside.
a if they are blanced,we put them aside.
b. if they are´nt balanced.
we should weigh the heavier two with any other two coins : now... if they balanced so the counterfeit coin is one of the two light coins,which will be easy to be find with the third weigh.....and if they are heavier,then we should weigh them one against one,,,and they heavier is the counterfeit coin.
2 for the 2nd five .we will do the same 2 against 2,leaving one coin aside:
a if they are balanced, so one of the two coins left aside is the counterfeit coin, which will be easy to find with the 3rd weigh.
b if they are not balanced....see above.

On 25.3.2016 at 2:13 AM, Mr.Dog said:
Alright, think i got it after fiddling with the unique number values for the letters.
Sol. abedcg = 5
( a=6 , b = 2 , c = 3 , d = 8 , e = 2, f = 1 , g = 4),
steps:
1. fedcg=abcdef 2. (fg)(bc)= (3)+(1)=4 3. (ab)cdef=0 > (g)cdef=0 > (gf)cde=0> 3cde=0
g=ab
4. Trial and error with abged = 2 , abgcd = 3 and 3cde = 0, easier when you know that g=ab.
abcdef = 0
bc = 1
abged = 2
abgcd = 3
fgbc = 4
abedcg = 5
afedcg = 6
abc = 7Amasing!!! you did find all the sequences from 0 to 7 ...but 5 is wrong..

13 hours ago, Mr.Dog said:
afhbg?
No..sorry...try again
Note that: abcdef= 0
On 20.3.2016 at 9:10 PM, wolfgang said:abcdef
bc
abged
abgcd
fgbc
.........?
afedcg
abcNote that: abcdef= 0

It seems to be hard....so let me give you a hint....which is:
5

abcdef
bc
abged
abgcd
fgbc
.........?
afedcg
abc 
I'm with dgreening  thought it was...
Hidden Content
Great!

dgreening and I have noticed the same thing:
Hidden Content
still not the right numbers

Tomaca Pipsuka

Hint....( think of....primary numbers)

I'll try
Hidden Content
no...they are wrong

Hello,,,who can give me the next three numbers :
74 , 34 , 14 , 73 ,13 , 92 , 32 , (?) , (?) , (?)

I think...two times

If the first digit is unique...so
There are alot of possible solutions , one of these is:
Andy........125
Bob.........213
Tim..........342
Fred........432
greg........513

Sorry...may I ask a question?.
Is it true for the digits in the middle ot at the end of the choosen number to be unique also?

the first position was 1:12 and the 2nd position was 3:24
and the difference between them is 2hours and 12 Min. 
with 5
e.g.
1st. 5
2nd 1
1st 5, 2 ( 5+2=7...he needs 8)
2nd 1,8 to prevent the 1st player to use it...( 1+8=9... the 2nd player needs 6 )
1st 5,2,6 to block the 15 for the 2nd...( 5+2+6=13... he needs 2)
2nd 1,8,2 blocking the 15 for the 1st.
1st 5,2,6,4 and wins because(5+6+4= 15)
The same will be true for all other numbers choosed by the 2nd. player and the 1st. player will win 
(You will answer me with" False" to 1+1=2)
The UTM will respond with " False "..which means that my expectation was true, and it responds with false .
There is no possiblity to respond with " True ", because 1+1=2
Secret message
in New Logic/Math Puzzles
Posted
e.g.( As the numbers are from 1 to 6 ) can be written like this: 1126621531323323123152261152311452242363622464