Roman numeral stick puzzle, 23/7=2
Walter Francis
answer, and after many days of trying to figure it out I give up!
His puzzle is this.. Move one 'stick' from the roman numerals to make
the equation true.
XXIII
----- = II
VII
By moving, that means you can make two II's an X or V, make it 22/8=2,
whatever.. I'm not sure if the roman numerals have to be in the proper
order (I had an example as I was working on it earlier, but it's late
and I'm tired..)
I appreciate any solution.. My brain is worn out.. :) My latest
attempt to figure it out went as such (and it might not wrap so well, so
bear with me..)
Numbers in parentheses indicate a 'stick' moving within that part of the
equation, such as making II a V. Numbers in square brackets indicate
adding a stick to that part of the equation, numbers in curly brackets
indicate removing a stick from that part. I went about creating any
possible numbers, and organizing them in this fashion in a hope to
'brute force' the answer, with no luck.. Here is what I came up with..
(18, 24, 26, 31, 63) [24, 27, 32] {14, 22}
-------------------------------- = (5, 10, 50) [3, 4, 6, 9, 11, 51] {1}
(12, 52) [8] {3, 6}
--
Walter Francis
http://wally.hplx.net Powered by RedHat 6.0
jon wild
: His puzzle is this.. Move one 'stick' from the roman numerals to make
: the equation true.
: XXIII
: ----- = II
: VII
The wording of the puzzle says to move a stick *from* the numerals, but
not *to* anywhere... take off one stick from the second X and you get:
XIIII
----- = II
VII
which will have one slightly wonky I (no wonkier than the Xs formed by
crossing a I with another I, explicitly allowed in the problem), but
it holds true. If you feel you absolutely must do *something* with the
stick, add it to the dividing line:
XIIII
------ = II
VII
I hope this satisfies your friend... cheers, jon
Jabuti
wa...@pop.uky.edu wrote:
XXII __
----- = II
VII
22/7 is appr. pi.
Jab
Sent via Deja.com http://www.deja.com/
Before you buy.
JCD
>A friend of mine teased me with a puzzle and refuses to give me the
>answer, and after many days of trying to figure it out I give up!
>His puzzle is this.. Move one 'stick' from the roman numerals to make
>the equation true.
>XXIII
>----- = II
> VII
>By moving, that means you can make two II's an X or V, make it 22/8=2,
>whatever.. I'm not sure if the roman numerals have to be in the proper
>order (I had an example as I was working on it earlier, but it's late
>and I'm tired..)
>I appreciate any solution.. My brain is worn out.. :) My latest
>attempt to figure it out went as such (and it might not wrap so well, so
>bear with me..)
I believe the answer yr friend have is a wrong answer. Many years ago,
people used to say that pi (3.14) was equal to 22/7.
22/7 = 3.14286, and pi=3.14159
My grandfather told me that a circle with a diameter of 7 has a perimeter
of 22. You can find this story in many old books, and I believe many
people in old centuries made this simple assumption.
The answer yr friend wants is maybe this:
XXII _
----- = II
VII
Rgds
JCD
JCD
Otto Edelenbosch
Walter Francis <wa...@pop.uky.edu> schreef in berichtnieuws
386886AE...@pop.uky.edu...
> A friend of mine teased me with a puzzle and refuses to give me the
> answer, and after many days of trying to figure it out I give up!
>
> His puzzle is this.. Move one 'stick' from the roman numerals to make
> the equation true.
>
> XXIII
> ----- = II
> VII
>
> By moving, that means you can make two II's an X or V, make it 22/8=2,
> whatever.. I'm not sure if the roman numerals have to be in the proper
> order (I had an example as I was working on it earlier, but it's late
> and I'm tired..)
>
> I appreciate any solution.. My brain is worn out.. :) My latest
> attempt to figure it out went as such (and it might not wrap so well, so
> bear with me..)
>
> equation, such as making II a V. Numbers in square brackets indicate
> adding a stick to that part of the equation, numbers in curly brackets
> indicate removing a stick from that part. I went about creating any
> possible numbers, and organizing them in this fashion in a hope to
> 'brute force' the answer, with no luck.. Here is what I came up with..
>
> (18, 24, 26, 31, 63) [24, 27, 32] {14, 22}
> -------------------------------- = (5, 10, 50) [3, 4, 6, 9, 11, 51] {1}
> (12, 52) [8] {3, 6}
>
> --
> Walter Francis
> http://wally.hplx.net Powered by RedHat 6.0
S P O I L E R below
Hi Walter,
All the answers already given are not true and by saying this i gave you the
answer.
Otto Edelenbosch
i4one...@cistron.nl (remove NOSPAM)
Tjeerd Kootstra
> XXIII
> ----- = II
> VII
>
XXII
---- <> II with the "=" covered by an I
VII
Ashok
>
>A friend of mine teased me with a puzzle and refuses to give me the
>answer, and after many days of trying to figure it out I give up!
>
>His puzzle is this.. Move one 'stick' from the roman numerals to make
>the equation true.
>
>----- = II
> VII
>
1. If two Roman numbers with a gap between them counts as multiplication,
this will work: move the middle I of the III in the numerator to the
right hand side, yielding
XXI I
----- = III
VII
2. An imperfect, but cute, answer is:
XXII _
----- = II
VII
22/7 is a decent approximation to pi.
Ashok
Drew Thomas
>22/7 = 3.14286, and pi=3.14159
No it's not! :)
It's 3.14159265358979323846264388!
--
Grats, Drew Thomas
"There are two kinds of people I never get involved with: serial killers and
people who spell 'thanks' with an x." -- Jonathan Creek
Hey! Think it's a hoax? Go to http://www.hoaxkill.com! If it's not there,
register it!
Don't do business with spammers!
Alan Quinn
>Walter Francis escreveu na mensagem <386886AE...@pop.uky.edu>...
>>A friend of mine teased me with a puzzle and refuses to give me the
>>answer, and after many days of trying to figure it out I give up!
>
>>His puzzle is this.. Move one 'stick' from the roman numerals to make
>>the equation true.
>
>>XXIII
>>----- = II
>> VII
>
>
>>By moving, that means you can make two II's an X or V, make it 22/8=2,
>>whatever.. I'm not sure if the roman numerals have to be in the proper
>>order (I had an example as I was working on it earlier, but it's late
>>and I'm tired..)
>
>>I appreciate any solution.. My brain is worn out.. :) My latest
>>attempt to figure it out went as such (and it might not wrap so well, so
>>bear with me..)
>
>
>I believe the answer yr friend have is a wrong answer. Many years ago,
>people used to say that pi (3.14) was equal to 22/7.
>
>22/7 = 3.14286, and pi=3.14159
Pi is approximately 3.14159265358979323846...
>
>My grandfather told me that a circle with a diameter of 7 has a perimeter
>of 22. You can find this story in many old books, and I believe many
>people in old centuries made this simple assumption.
>
>The answer yr friend wants is maybe this:
>
>XXII _
>----- = II
> VII
>
>Rgds
>JCD
>
>
girton2
Signed,
Dumb Founded
Drew Thomas wrote in message <+R3zLPAn...@michthom.demon.co.uk>...
>
>>22/7 = 3.14286, and pi=3.14159
Rich Grise
>
> A friend of mine teased me with a puzzle and refuses to give me the
> answer, and after many days of trying to figure it out I give up!
>
> His puzzle is this.. Move one 'stick' from the roman numerals to make
> the equation true.
>
> XXIII
> ----- = II
> VII
XXIII
----- á I
VII
(where the "á" is "Is Not Equal To")
Cheers!
Rich
Rich Grise
light-years, what's the error if you calculate the circumference using
the two following approximations of pi?
3.14159
3.14159265358979323846264388
Happy Odometer Year!
Rich
girton2
>Ob Puzzle: Assuming the radius of the known Universe to be 12 billion
>light-years, what's the error if you calculate the circumference using
>the two following approximations of pi?
>
>3.14159
>
>3.14159265358979323846264388
0.00003184307751886155172656 light years, (the 1/x being
31403.9997989412874233462223411074)
or
186781812.43007612851627608576 Miles
Signed,
Dumb Founded
girton2
girton2 wrote in message <84lnad$6qd$1...@oak.prod.itd.earthlink.net>...
Rich Grise
0.00003184307751886155172656 light years
----------------------------------------
12,000,000,000 light years
Which is about an error of 2.5 parts in 10e16.
I think I'll stick with 3.14159 for all practical purposes.
Thanks!
Rich
girton2
However, it's good to know that you have a grasp of a calculator.
Signed,
Imus Be Stoopid
Rich Grise wrote in message <386F08...@entheosengineering.com>...
Public User
>
> A friend of mine teased me with a puzzle and refuses to give me the
> answer, and after many days of trying to figure it out I give up!
>
> His puzzle is this.. Move one 'stick' from the roman numerals to make
> the equation true.
>
> XXIII
> ----- = II
> VII
>
Move the last stick from XXIII and put it horizontally on top of the II
on the right hand side.
This gives the approximation:
22 / 7 = Pi
Yours truly,
Will
anilpra...@gmail.com
----- > II 23/7=2 3.2=2 3.2>2
VII