|
Post by Apolonio on Nov 30, 2004 20:57:52 GMT -5
I won't post it unless I know it's an interesting topic to some, I do want to post one if any of you find it interesting.
Rather than explain the problem at first, I'll post the complete problem with an answer and then you gals and guys can figure out "code breakers" break the code "formula" to the math trick.
Picked up this math trick from an 1890's math book many years ago and never forgot the trick, Never seen it or even spoken of in my lifetime anywhere else since, It's the best disguised math trick ever.
OK, You up for code breaking? Let me know and I'll post the problem with an answer and let's see if you can break the code!
HH, Paul (Ca)
|
|
TONKA
FOAP (Father of All Posters)
Posts: 1,010
|
Post by TONKA on Nov 30, 2004 21:05:59 GMT -5
I doubt that I'll be able to solve the problem but I, for one, am interesting in trying! ;D
|
|
|
Post by Rudy on Nov 30, 2004 21:42:28 GMT -5
I'll ditto Tonka. ;D In the meantime, there were two sheep herders. The first sheep herder asked the second one to sell him a sheep so that he would have twice as many sheep as the other. The second herder replied that it would be better if the first one sold him a sheep and then they would have the same number of sheeps. How many sheep does each one have?
|
|
|
Post by Apolonio on Dec 1, 2004 16:30:04 GMT -5
Hello TONKA and Rudy, OK, I'll post the problem in a day or on Friday. I'm so sure you and several of the others will enjoy this trick. Rudy, Thanks for posting your math problem, This will tease several of the others HH, Paul
|
|