Advertisement
 

Membership

 

Advertisement

 

Search

 
Forums \ General forum \ The riddle thread
 1, 2, 3 ... 13, 14, 15, 16 

3 months ago
#261

The_annoying_Devil

  3
2008-08-12 00:56:02, Saline wrote: In addition, there's another point below this where walking one mile east will take you TWICE around the Earth, and you'll still end up at the same spot. There's yet ANOTHER point where walking east will take you three times around. In fact, for every integer N there exists a point where walking East will take you N times around the Earth, and back to where you started.


Uuh, I didnt consider this. I had a bit stomachache because my one and only ring includes only a (fairly large but) finite set of "places". But taking it down to "N times around the Earth" is quite a better argument for infinity.
Top of the page   Quote  

3 months ago
#262

tornado

  1
lol so bad just lick this OWN
Top of the page   Quote  

3 months ago
#263

The_annoying_Devil

  3
A much easier version of the so called Lucifer riddle


'I have 3 children. Their ages multiplied gives 72. Sum up their ages to get my housenumber, but that won't help you. If I leave home the oldest one take care of the two others. How old are my children?'
Top of the page   Quote  

3 months ago
#264

Chairman.Meow

  3
The_annoying_Devil:

#261
Actually your "one and only" ring contains an (uncountably) infinite set of solutions. Each of Saline's rings (of which there are a countably infinite number) contains an (uncountably) infinite set of solutions. As it turns out, though, you both had the "same number" of solutions!

#263
a) You need to explain this sentence in your "Lucifer" riddle, as it makes no sense:
Gauss get the product and Euler the sum of the two numbers.


b) The only real information you give about the 3 children's ages is that they multiply to 72 (= 2*2*2*3*3). There are multiple ways to multiply 3 numbers together to get 72. If you had said 385 (= 5*7*11) or some other multiple of 3 primes, then there would be a unique solution.
Top of the page   Quote  

3 months ago
#265

Saline

  3
I agree with #264: the Gauss and Euler problem is very poorly worded/written. I think we've sadly missed some critical information due to grammatical mistakes. Here's my interpretation of the riddle:

Lucifer is thinking of two numbers x and y between 1 and 100. To Gauss he gives the product x*y, and to Lucifer the sum x + y. Gauss and Euler then converse:

Gauss: I don't know what the two numbers are.
Euler: I don't know what they are either, but I knew that you didn't know the two numbers.
Gauss: Now I know the two numbers.
Euler: I know them now as well.

I get the feeling this isn't exactly correct. Can you double-check your source?

-----------------------
The other riddle also was misworded--the poster took the riddle from somewhere else but missed the critical part of the riddle that gives the information required to solve it. Here's the real, solvable version of the 3 children's ages riddle:


A man told a census taker that he had 3 children. When asked their ages, he replied. "The product of their ages is 72." The census taker said, "That's not enough information." The man said, "The sum of their ages is the same as my house number." The census taker looked at the house number and said "I still can't tell." The man replied, "Sorry, I forgot to tell you that my oldest likes strawberry pudding." The census-taker then wrote down the ages of the 3 children. How old were they?
Top of the page   Quote  

3 months ago
#266

Saline

  3
Here's my next riddle:

Bill has two girlfriends, Hillary and Monica. Monica lives in the East of a city, and Hillary lives in the West of the same city. Once every morning at a random time, Bill arrives at the train station at the center of the city. A train leaves for the East every 10 minutes, and a train leaves for the West every 10 minutes — Bill chooses whichever train arrives first. On average, could Bill end up with one girl more often than the other? If so, how many times more often? Why?
Top of the page   Quote  

3 months ago
#267

Chairman.Meow

  3
2008-07-30 17:50:03, EffOrt wrote:
The set of numbers {9, 99, 999, 9999, ...} has some interesting properties. One of these has to do with factorization. Take any number n that isn't divisible by 2 or by 5. You will be able to find at least one number in the set that is divisible by n. Furthermore, you won't need to look beyond the first n numbers in the set.

Prove it.


I'm having a lot of trouble with this one. Here's where I'm at.

First of all, some preliminary stuff & terminology.
- Saying that a number "isn't divisible by either 2 or 5" is the same as saying it's "relatively prime" to 10, which is shorter to write.
- A rational number is a number of the form p/q, where p and q are integers. We are only interested in the case where p and q are positive.
- Every rational number has an (infinite) repeating decimal expansion. (In the case of finite decimals, we can rewrite them to fit our definition by decreasing the last decimal digit and attaching infinite 9's to the end. For example 1/2 = 0.5 becomes 0.499999... and thus has an infinite repeating decimal expansion.) The decimal expansion of a rational number consists of a possible non-repeating part followed by a repeating part.
- A rational number is said to have an immediate-repeating decimal expansion iff it has no non-repeating part. (4/7 = 0.571428571428..., for example). Decimal expansions with non-repeating parts are called delayed-repeating (4/15 = 0.266666..., for example).

OK, I can prove that the following 2 statements (which are contrapositives of each other) imply the first proposition. The problem is that I can't prove either of these!

(a) If n is relatively prime to 10, then 1/n has an immediate-repeating decimal expansion.
(b) If n has a delayed-repeating decimal expansion, then n is not relatively prime to 10.

Proof that (a) implies the first statement: Suppose (a) holds, and let n be any integer relatively prime to 10. Then 1/n has an immediate-repeating decimal expansion. Let (A1)(A2)...(Ak) be the repeating part (of length k). (By the way if anybody can tell me how to do subscripts I could make this look ALOT prettier! For now I'm gonna use commas to separate the "number" and "decimal" parts)We have

_______1/n = 0,(A1)(A2)...(Ak)(A1)(A2)...(Ak) . . . and
(10^k)*(1/n) = (A1)(A2)...(Ak),(A1)(A2)...(Ak)(A1)(A2)...(Ak) . . .

Subtracting the second equation from the first we obtain

(10^k - 1)*(1/n) = (A1)(A2)...(Ak)

Thus we have 10^k - 1 equal to n times an integer; in other words, n divides 10^k - 1. But 10^k - 1 = 99...9 (k 9's)!!! This completes the proof.

Now, if we has (a) and we could show that the length of the non-repeating part of 1/n never exceeds n then we would also have the second statement.

Anybody have any ideas on how to prove (a) or (b)? Anybody still care about this "riddle"???
Top of the page   Quote  

3 months ago
#268

Saline

  3
Chairman, I'll take a look at that in a second. I just stumbled upon another very easy (simple algebra) riddle with a rather amusing solution:

The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father?
Top of the page   Quote  

3 months ago
#269

Chairman.Meow

  3
#265


A man told a census taker that he had 3 children. When asked their ages, he replied. "The product of their ages is 72." The census taker said, "That's not enough information." The man said, "The sum of their ages is the same as my house number." The census taker looked at the house number and said "I still can't tell." The man replied, "Sorry, I forgot to tell you that my oldest likes strawberry pudding." The census-taker then wrote down the ages of the 3 children. How old were they?


There still isn't enough info! The only real information we've been given is that the ages of the children multiply to 72.
Top of the page   Quote  

3 months ago
#270

Chairman.Meow

  3
The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father?


Top of the page   Quote  

3 months ago
#271

Saline

  3
Chairman: I think you did your algebra wrong.


With regards to the {9, 99, 999} problem, I came across a very elegant solution. I didn't come up with this (it's been a long time since I had any group theory) but when you follow the logic it's pretty awesome!
Top of the page   Quote  

3 months ago
#272

Chairman.Meow

  3
2008-08-12 03:15:56, Saline wrote:
Here's my next riddle:

Bill has two girlfriends, Hillary and Monica. Monica lives in the East of a city, and Hillary lives in the West of the same city. Once every morning at a random time, Bill arrives at the train station at the center of the city. A train leaves for the East every 10 minutes, and a train leaves for the West every 10 minutes — Bill chooses whichever train arrives first. On average, could Bill end up with one girl more often than the other? If so, how many times more often? Why?


Top of the page   Quote  

3 months ago
#273

Chairman.Meow

  3
Chairman: I think you did your algebra wrong.




With regards to the {9, 99, 999} problem, I came across a very elegant solution. I didn't come up with this (it's been a long time since I had any group theory) but when you follow the logic it's pretty awesome!

That is quite elegant! Though it requires no group theory - only number theory.
Top of the page   Quote  

3 months ago
#274

Saline

  3
Right, but your answer is correct. The question asked how it was possible for Billy to end up with one girl more often than the other--the answer is if the average waiting time for the first train is less than the average waiting time for the second, which is what happens when they arrive close to each other!
Top of the page   Quote  

3 months ago
#275

Chairman.Meow

  3
Still don't get the census one . . . is it just meant to confuse?
Top of the page   Quote  

3 months ago
#276

Saline

  3
What you're missing is that the census taker didn't have enough information given the first two statements.

True, knowing that the product of the three children's ages are 72, we don't have enough information. There are quite a few options for the ages:

72 + 1 + 1 = 74
36 + 2 + 1 = 39
24 + 3 + 1 = 28
18 + 4 + 1 = 23
18 + 2 + 2 = 22
12 + 3 + 2 = 17
12 + 6 + 1 = 19
9 + 8 + 1 = 18
9 + 4 + 2 = 15
8 + 3 + 3 = 14
6 + 6 + 2 = 14
6 + 4 + 3 = 13

Now, remember that the census taker knew the house number, but he still didn't have enough information. This means that the sum of the children's age was not unique from the list!--if it was, he would know their ages. Therefore we can remove from the list all instances where the sum of the ages is unique, and we get the following list:

8 + 3 + 3 = 14
6 + 6 + 2 = 14

Note that now we need one more piece of information to narrow the results down. Sure enough, we have one more clue--the fact that the man has an eldest child. The second case has no eldest child--only the first does. Thus, the children's ages are 8, 3, and 3.
Top of the page   Quote  

3 months ago
#277

Beats_MC

  1
Saline.. Product is multiplication. Sum is Addition.
Top of the page   Quote  

3 months ago
#278

The_annoying_Devil

  3
2008-08-12 03:13:34, Saline wrote:
Gauss: I don't know what the two numbers are.
Euler: I don't know what they are either, but I knew that you didn't know the two numbers.
Gauss: Now I know the two numbers.
Euler: I know them now as well.


The source and nearly every other source forgot to say, that both, product and sum are < 100 aswell. So we have x,y,x*y,x+y < 100.
edit: some other sources don't allow square numbers.
Top of the page   Quote  

3 months ago
#279

Saline

  3
I'm aware of that. All of the numbers I showed above each have product 72. It's the sums we have to look at--the sum of the children's ages equaled the house number--it's the sums that interest us.
Top of the page   Quote  

3 months ago
#280

The_annoying_Devil

  3
What are you talking about? The riddle with the ages is solved. 8-3-3 is the correct answere. I was refering to that Lucifer riddle.

Nevermind, try a next one.
A natural number consists only of 1 and 0 (for example 100110011 or 10 decimal! notation). In our case there are exactly 300 1 in that number and an unknown amount of 0. The number shall be a square number.
Prove that there is no number that will fulfill the conditions.
Hint:
Top of the page   Quote  

 1, 2, 3 ... 13, 14, 15, 16 

Forum post
  • Do to others what you would have them do to you.
  • Do not judge so that you will not be judged.

    » Read posting rules in detail

    Reminder: IP-addresses for each comment is saved. Offensive posts will be censored. Repeated offense will result in a warning and/or a temporal lockdown of the account.
  • The riddle thread \ Reply
    To submit a post, you have to be a registred member. Register an account below the title 'Membership' in the left menu.

    » Top of the page