The worlds first riddle!

I think you've shown great cultural sensitivity, Tryagain. You didn't offend Democrats. Nor did you offend us turtles! I mean, in most illustrations of how the Fibonacci sequence plays out in real life they will use rabbits. You've steered clear of that kind of outrageous lapine triumphalism.

How many offspring would a pair of turtle hatchlings have in their first year?
None! We have 100-year lifespans, so there's no rush at all.

Could we get back to the riddle? I love this riddle and would like to hear the answer. Can we live forever? We dont know until we die. But i bet it is a metaphor not a straight out philosophy question.

I love this. I think it is a metaphor rather than a philosophy question? If a philosophy question i would go with Can we live forever? We will only know when we die. As a metaphor, i hope it is something cool like ???

Mark continues his winning streak…With a bonus for his wit!

“hetero hamsters: Per Leonardo of Pisa, 233 pairs at the end of the 12th month.
homo hamsters: Only 1 pair, but their enclosure is decorated very fashionably.”

Let R(n) be the number of rabbit pairs at the end of month n. Clearly, R(0) = 1
and R(1) = 1. For every n > 1, the number of rabbit pairs, R(n), is equal to the
number of pairs at the end of month n − 1, R(n − 1), plus the number of rabbit
pairs born at the end of month n. According to the problem’s assumptions, the
number of newborns is equal to R(n − 2), the number of rabbit pairs at the end of
month n − 2. Thus, we have the recurrence relation
R(n) = R(n − 1) + R(n − 2) for n > 1, R(0) = 1, R(1) = 1.
The following table gives the values of the first 13 terms of the sequence, called the
Fibonacci numbers, defined by this recurrence relation:
n 0 1 2 3 4 5 6 7 8 9 10 11 12
R(n) 1 1 2 3 5 8 13 21 34 55 89 144 233

My thanks to Fibonacci.
Acknowledgments to Kolyo.

Hi Kolyo,
I once had a pet turtle that I used to take to the beach, till one day it suddenly falls over dead in the parking lot. I ran to call the vet…

A passing cop yells: "Hey buddy, you can't just leave that lyin' there."
To which I replied: "That's not a lion officer, that's a turtle!" It wasn’t the time or place for outrageous Chelonia triumphalism.

If a turtle loses its shell...is it naked or homeless?


Здравствуйте Stan, Рад Вас видеть kак поживаешь?

“I think it is a metaphor rather than a philosophy question”

Could it be a Metaphor in analytic philosophy!

If we take interactionist theory which asserts that at the heart of a metaphor is the interaction between its two subject terms, where the interaction provides the condition for a meaning which neither of the subject terms possesses independently of the metaphorical context.

Whereas analytic philosophy examines metaphor within the philosophy of language, continental philosophy assigns much wider significance to metaphor, which has proven to be extremely important for this thinking because it is the process of conceptual borrowing or reassignment which revises our perception of the world.

Only when a person has died will they know the true answer to the question, ‘Will I live forever?’

Ps. welcome to A2K, I hope you enjoy your stay.


A vehicle odometer can display any six-digit combination from 000,000 to 999,999, inclusive.

If it runs through its entire range, how many such combinations will have at least one digit 1 in them?

What is the total?

odometer: There are 10^6 total combinations. There are 9^6 combinations using the digits 0 and 2-9 (no 1s). Therefore, there are 10^6 - 9^6 = 468559 combinations with at least one 1.

Mark:
Odometer: There are 10^6 total combinations. There are 9^6 combinations using the digits 0 and 2-9 (no 1s). Therefore, there are 10^6 - 9^6 = 468559 combinations with at least one 1.

Indeed…
The answer to the first: 468,559 and ….
Part two: 600,000, respectively.

To answer the second question, one can simply note that each of the 10 digits will appear in all the readings the same number of times. Therefore the total number of times a 1 will appear can be obtained as one tenth of the total number of all the digits in all the readings: 0.1(6 • 10^6) =600,000.


There are 12 very smart Democrat A2K members in a Southern jail on a trumped up charge of ordering a super-sized soda! To get rid of them, the evil Republican Sheriff comes up with the following test.

He will put a hat, either blue or white, on the head of each of these prisoners. There will be at least one hat of each color, and the prisoners will be informed about this fact.

They will be able to see everyone else’s hat but their own; there will be no communications of any kind among the prisoners.

The Sheriff will line up the prisoners every 5 minutes starting at 9:05 pm and ending at 9:55 pm. To pass the test, all the prisoners with a blue hat and only those prisoners will have to step forward during the same line up. If they do, all the prisoners will be freed; otherwise they will be forced to support a bill to make Richard Milhous Nixon a saint.

How can the innocent A2K prisoners pass the test?


Please help; reports suggest they are being force fed Hominy grits and Red-eye gravy!

As long as all of the smart prisoners who are democrats can count their 123's they can save themselves. Every five minutes, starting at 9:05 pm the prisoners will form a line as commanded by the sheriff for a total number of eleven times. The eleventh time, at 9:55 pm being the last time a line can be formed where only the democrats with a blue hat must step forward.

Assuming that the prisoners have the freedom to line behind anyone, seeing that it isn't specified whether the sheriff controls what order they are to line up, at 9:05 pm, the prisoners line up and memorize the person's feature in front of them as well as count and remember all the red hats they can see in front of them (F). At 9:10 pm the prisoners line up behind the same person they were with at 9:05 pm and face the opposite way (so as to count the number of red hats behind them, "behind" relative to the position that they were at 9:05 pm). Each of the prisoners count and remember the number of hats "behind" them (B). In the next five interval, the prisoners total the number of red hats in front of them and "behind" them (F+B=T), then follow the logic:
If T was even, and if F and B are either both even or are both odd, then the wise democrat reasoning this logic would know the hat being worn is blue. Otherwise the wise democrat would know the hat being worn is red.

If T was odd, and if F and B are either both even or are both odd, then the wise democrat reasoning this logic would know the hat being worn is red. Otherwise the wise democrat would know the hat being worn is blue.

At 9:15 pm those who know to have a blue hat will be the only ones to step forward while the red hatters will stay in their spot, and voila they are set free and released 30 minutes earlier than the scheduled deadline.

I'm not following this. It seems that all each prisoner knows is how many blue hats are worn by other prisoners and how many red (white) hats are worn by other prisoners. Since a prisoner doesn't know his own hat color, he can't do a parity check.

However, your method gave me an idea:
First lineup: Send them out one at a time. Starting with the second prisoner, a prisoner stands to the left of the prisoner who preceded him if that prisoner is wearing blue, otherwise he stands at the far right of the line. All but the last prisoner knows his hat color.
Second lineup: Send the formerly last prisoner out first. The second prisoner follows the rule from the first lineup. All the rest go to the far right of the line. Now all prisoners know their hat color. Once they're all lined up for the second time, all that are wearing blue step forward.

The commute home brought some clarity to the problem. This is an oldie in a different setting. If there are N prisoners with blue hats, then they will all step forward at the Nth lineup.

Assume just one: He'll be the only one to see zero blue hats, so he'll step forward.
Assume two: Since nobody stepped forward in the first lineup, everybody knows that there are at least two. Whoever sees just one will step forward.
And so on...

Yes, that would work. I forgot to acknowledge the parity check. Possibly in the first lineup the person standing last in line who can see everyone but them-self will cough once if he sees an even number of red or not cough at all if the number of red hats are odd, and let (F) be the number of red hats he can see. Then each following person can keeps track of the number of red hats known to have been saved from behind (B) in the second line formation, and counts the number of red hats in front (F).
But then again I'd think coughing would be considered cheating.
The simple process of elimination that you mentioned is the straightforward method without any "cheating" per se.
I over thought this one.

Nice try Ben – but no cigar this time. However you were on the right track and did point out the time factor and added another color as well as giving Mark a clue as to the eventual answer – so out of the kindness of my heart… I award you points de néant.

Mark, in the interest of highway safety, may I respectfully recommend you, “Keep your eyes on the road and your feet on the peddle, and the peddle to the metal”

As Mark has pointed out…

Suppose only one of the hats is blue. Then the prisoner wearing it would see only white hats on the other prisoners. Since he knows that there is at least one blue hat, it must be his. Each of the other prisoners sees one blue hat and can make no definite conclusion about the color of their hats.

So the prisoner who sees only white hats is the only one who steps forward during the first line up, which brings the freedom to all of them.

Suppose there are two blue hats. Then no prisoner steps forward during the first line up because none of them can be certain about the color of his hat. But during the second line up both prisoners who see one blue hat can step forward.

Indeed, since no prisoner stepped forward during the first line up, they know that there are at least two blue hats. If a prisoner sees just one blue hat, he can conclude that the second blue hat is his. All the prisoners who see two blue hats cannot be certain about the color of their hats and therefore will stay in line.

In general, if there are k blue hats, 1 ≤ k ≤ 11, no prisoner will step forward during the first k − 1 line ups. But on the kth line up, each of the prisoners who sees k − 1 blue hats will be able to step forward by the same reasoning as above:

He knows that there are at least k blue hats since the previous line up did not stop the test. If he sees k − 1 on the other prisoners, then there are exactly k blue hats, one of which is his. At the same time, all the n − k prisoners with white hats should—and given their smartness will— stay in line because they cannot deduce the color of their hats.

The smarts can win freedom after all….. up yours Nixon!

I would like at this juncture to give praise to Missy; the only one who cared enough about the suffering to have baked an apple pie for them. Unfortunately the evil Republican seized it and ate it hisself - and mighty fine it tasted too!

Mark is right to assert, “This is an oldie in a different setting…” there is a dearth of humor amongst the multitude; replaced by murder, death, kill exchanges. Therefore it is refreshing to note that no threats are attached to this first time asked conundrum…


What 5-digit number has the following features?

If we put the numeral 1 at the beginning, we get a number three times smaller than if we put the numeral 1 at the end of the number.

And if anyone has the temerity to get this right, “There are certain things that I can do, with hairspray, Nair and superglue, and sooner or later, you've got to sleep." N'est-ce pas!

Question, if I may. I have not done any of the difficult puzzles, but I do want to give this last one a try.

I think I have figured out two of the numbers, but there is so much thinking involved, I want to check before going on.

Are questions submitted to you by PM allowed. The question will be a single question...and I am content to simply get a "you are on the correct track"...just want to know if it is allowed.

Never mind...I've got the answer. How do I submit it?

I got it also....

Congratulations, SM. I submitted mine by PM. I am guessing that is the correct procedure.

Can I submit it here? or should I send a PM?

I will submit it here since Frank already sent a PM anyways...

14257 x 3 = 42771...

If I understood the parameters correctly...

I CANNOT IMAGINE YOU SHOULD SUBMIT IT HERE...you will spoil it for others trying to solve it.

Send a PM.

I do not think you did understand the parameters, SM.

The puzzle called for a 5 digit number that if you added a one...

Yours is only 4 digits.

Although there are two ways to read the problem. I solved it as a five digit number with a one in front of the 5 and a 1 at the end of the 5.