Tommie Hu$tle
Member
Anyone?
a puzzle;
(space)= - or .
fccjgbfcg egjgbeebbibeejgbgfcghg egbiaabbgdeie jfcc eiejgb bibijgbgfc
a puzzle;
(space)= - or .
fccjgbfcg egjgbeebbibeejgbgfcghg egbiaabbgdeie jfcc eiejgb bibijgbgfc
A puzzle
- Thread starter Tommie Hu$tle
- Start date
- Status
Particle Physicist
between a quark and a baryon
what ? ?
Tommie Hu$tle
Member
quadriplegicjon said:
That's what I'm saying. It's a puzzle that has a solution but, I'll be dammed if I can figure it out. I was hoping someone smarter than me could figure it out.
That's all I got to go on.
Manabanana
Member
Tommie Hu$tle said:
It's a math problem. The letters don't go above "J", the 9th letter so you can turn all the numbers into letters. The confusing part is the spaces. Either it's all subtraction, or you have to get lucky with decimal placement. At least that's my guess.
Edit: J is the 10th letter...MANA USE FINGERS COUNT LETTER. I have no idea.
Tommie Hu$tle
Member
What I do know is that it is a link to a website.
Particle Physicist
between a quark and a baryon
where did you see this puzzle?
BugCatcher
Member
J is the 10th letter, but if A = 0 then J = 9.
Tommie Hu$tle
Member
That gives us
522961523 - 4696144118144961652676 - 4618001163484 - 9522 - 484961 - 1818961652
or
522961523 . 4696144118144961652676 . 4618001163484 . 9522 . 484961 . 1818961652
522961523 - 4696144118144961652676 - 4618001163484 - 9522 - 484961 - 1818961652
or
522961523 . 4696144118144961652676 . 4618001163484 . 9522 . 484961 . 1818961652
Tommie Hu$tle
Member
slayn said:
That's kinda what I'm thinking but, I can't seem to tell what is what.
I'm too tired to go after this right now, but since the letter space is a-j, you might be looking at an Affine cipher with modulus 10.
By the way -- what was the prior content of his blog, and why are you so interested in this puzzle? (PM me if it's not appropriate for general consumption.)
By the way -- what was the prior content of his blog, and why are you so interested in this puzzle? (PM me if it's not appropriate for general consumption.)
As a side puzzle, my brother once sent me this puzzle claiming that Einstein couldn't solve it or something. I didn't believe him, since I solved it (and I'm definitely no Einstein):
The Penny Puzzle.
You have twelve pennies and a set of scales. One of the pennies is a 'bad' penny and is a different weight to the other pennies. What is the least number of weighs (uses of the scales) required to determine which penny is the 'bad' penny (and whether it is heavier or lighter)?
Short answer:
The Penny Puzzle.
You have twelve pennies and a set of scales. One of the pennies is a 'bad' penny and is a different weight to the other pennies. What is the least number of weighs (uses of the scales) required to determine which penny is the 'bad' penny (and whether it is heavier or lighter)?
Short answer:
I believe it's 3, but the REAL puzzle is whether you can figure out how it is done!
mrkgoo said:
not too difficult, but maybe cause you do crap like this (searches) all the time in Comp Sci. Step 1. make two batches of six, weigh em. The bad penny is in the heavier one
Step 2. seperate bad penny bunch into two piles of 3. Weigh em. Bad penny is in the heavier one
Step 3. Choose any two of the remaining 3, if they don't weigh the same you've found the penny, if they do weigh the same then the penny you didn't weigh is the bad one.
Step 2. seperate bad penny bunch into two piles of 3. Weigh em. Bad penny is in the heavier one
Step 3. Choose any two of the remaining 3, if they don't weigh the same you've found the penny, if they do weigh the same then the penny you didn't weigh is the bad one.
Particle Physicist
between a quark and a baryon
Azih said:
how do you know that bad penny is the heavier one? he said it could be heavier or lighter
Baron Aloha
A Shining Example
12 pennies.... split them into two stacks of 6 pennies each (stack 1 and stack 2).
Split stack 1 into two 3 penny stacks (stack 3 and stack 4).
Weigh stack 3 and stack 4.
If stack 3 and stack 4 weigh the same amount then you know the bad penny is in stack 2. Otherwise the bad penny was in stack 1.
Split the bad stack into 3 stacks of 2 pennies each (stack A, B, & C).
Weight stack A & B. If they weigh the same the bad penny is in stack C. If they don't weigh the same then weigh stacks A & C. If stacks A & C weigh the same the bad penny is in stack B. If they don't the bad penny is in stack A.
Choose 1 penny at random from stack A, B, or C (whichever stack had the bad penny). Weigh it against any other penny besides the one it was paired with. If it weighs the same then the penny you didn't choose is the bad penny. If it weighs more or less then the penny you chose is the bad penny.
At most it should take a total of 4 weighs on the scale to find the bad penny.
Split stack 1 into two 3 penny stacks (stack 3 and stack 4).
Weigh stack 3 and stack 4.
If stack 3 and stack 4 weigh the same amount then you know the bad penny is in stack 2. Otherwise the bad penny was in stack 1.
Split the bad stack into 3 stacks of 2 pennies each (stack A, B, & C).
Weight stack A & B. If they weigh the same the bad penny is in stack C. If they don't weigh the same then weigh stacks A & C. If stacks A & C weigh the same the bad penny is in stack B. If they don't the bad penny is in stack A.
Choose 1 penny at random from stack A, B, or C (whichever stack had the bad penny). Weigh it against any other penny besides the one it was paired with. If it weighs the same then the penny you didn't choose is the bad penny. If it weighs more or less then the penny you chose is the bad penny.
At most it should take a total of 4 weighs on the scale to find the bad penny.
CaptainABAB
Member
I used 4 stacks of 3 pennies but I still require 4 weighs depending on which one it is. I can't get it down to 3 weighs.
In that case, lets say its heavier.mrkgoo said:
Divide in 3 stacks of 4. Weigh two of them, the heavier stack will have the heavy penny, if they weigh the same the remaining stack has the heavy penny. Now divide stack in two, weigh stacks and heavier stack has heavy penny. Weigh last 2 remaining pennies to see which one is heavier.
bad penny puzzle, does it go something like this?
divide into 3 stacks.
t1=[s1, s2, s3, s4] t2=[s5, s6, s7, s8] t3=[s9, s10, s11, s12]
then weigh.
if t1= t2
make u1=[s1, s9] u2=[s10, s11]
if u1 = u2 then s12 is the "bad" penny.
of course this formula differs depending on what the initial t1 and t2 weighing are. if one is greater or less then the other then your second weighing would consist of something like:
u1=[s1, s2, s5] u2=[s3, c6, c12]
which then would determine what the final two pennies you would weigh. and i don't want to type out those combinations
man, that sounds confusing.
t1=[s1, s2, s3, s4] t2=[s5, s6, s7, s8] t3=[s9, s10, s11, s12]
then weigh.
if t1= t2
make u1=[s1, s9] u2=[s10, s11]
if u1 = u2 then s12 is the "bad" penny.
of course this formula differs depending on what the initial t1 and t2 weighing are. if one is greater or less then the other then your second weighing would consist of something like:
u1=[s1, s2, s5] u2=[s3, c6, c12]
which then would determine what the final two pennies you would weigh. and i don't want to type out those combinations
man, that sounds confusing.
alejob said:
No, no, I meant that my answer will determine whether that it is heavier or lighter within the three weighings, not that it is known from the beginning.
somnific said:
However, if u1 < u2, you don't know if s9 is the bad penny and lighter, or if it's s10 or s11 that's heavier.
somnific said:
Actually, I take what I just said back that part works out ok, I think. I can't really translate the last part though. What's c6 and c12?
edit: on closer inspection, I think that works out (maybe have a few others check it for accuracy), but it's similar but not identical to my answer. Congrats!
- Status