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.
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.
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.)
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:
I believe it's 3, but the REAL puzzle is whether you can figure out how it is done!
I think it definitely might be an Affine cipher. Look at how often the first six letters show up compared to the others...I think there's something there.
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:
I believe it's 3, but the REAL puzzle is whether you can figure out how it is done!
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.
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.
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.
I think you have to know if the penny is heavier or lighter. I heard a similar puzzle but there were 9 pennies and 1 was heavier. You get 2 tries on the scale to see which penny is the heavier one.
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.
That's good...but it still might take 4 weighs... I'm pretty sure my answer gets it in 3 weighs, regardless, and you Know whether the bad penny is heavier or lighter.
That's good...but it still might take 4 weighs... I'm pretty sure my answer gets it in 3 weighs, regardless, and you Know whether the bad penny is heavier or lighter.
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.
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
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.
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.
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
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
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!
