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Forums \ General forum \ The riddle thread
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3 months ago
#281

Saline

   3
No, no...I was talking to Beats_MC, who implied that I got my multiplication and addition mixed up.
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3 months ago
#282

Saline

   3
Annoying_Devil: I think I can take a crack at the Einstein/Gauss riddle, although I won't finish it here, but I can get people on the right track (I believe) if they want to solve it. Also, I looked up the wording of the riddle online, and it's also necessary that x and y be two different numbers greater than 1 and less than 100. The wording of the riddle, to clarify, goes like this:

(Einstein has product, Gauss has sum)

Einstein says:"It seems that I can not tell which are the two numbers"
Gauss says:"I knew you couldn't"
Einstein says:"If so I know now which they are "
Gauss says:"If so I know them too"

We need to make two lists: a sum list and a product list. Let's begin with a simple list for both:

P: 2, 3, 4, 5, 6, ...
S: 2, 3, 4, 5, 6, ...


Next, remember that the numbers must be unique and > 1. Therefore, the numbers 1+1=2, 1+2=3, 1+3/2+2 = 4 are taken out of the sum list. From he P list we can use the same logic: 1*2=2, 1*3=3, 2*2=4, 5*1=5, 3*2=6, 3*3=9, etc.
S: 5, 6, 7, 8, ...
P: 7, 8, 10, etc.

Next: Einstein doesn't immediately know the answer.
Let's find numbers where Einstein would immediately know the answer. These are numbers with two distinct factors, also known as products of two (distinct) primes. Just to take a few from the beginning:
primes (easy): 1, 2, 3, 5, 7, 9, 11, 13, 17, 19...
2*3=6, 2*5=10, 2*7=14, 2*11=22, ...
3*5=15, 3*7=21, 3*11=33...
5*7=35 ...

What we've removed from this list is:
6, 10, 14, 15, 21, 22, 33, 35.
Our P list now looks like:
P: 8, 12, 16, 18, 20...
In addition, we can remove the sums of those pairs of primes (since otherwise Einstein would know the product, and Gauss knew he didn't know):
2+3=5, 2+5=7, 2+7=9, 2+11=13, 2+13=15, 2+17=19, ...
3+3=6, 3+5=8, 3+7=10, 3+11=14, 3+13=16, 3+17=20, ...
5+7=12, 5+11=16, 5+13=18, 5+17=22, ...
Our sum list now looks like this:
S: ...we're already above 18

That's a start...I think you could continue and eliminate even more numbers. It's a tough one!
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3 months ago
#283

Chairman.Meow

   3
#280 So is this a number written in base-10 that contains only 0's and 1's or is it a number written in binary?
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3 months ago
#284

user_error

   4
"#250 You're a jerk! Though the math isn't hard: 7^4."

Close, it would be 7^4 + 1 (don't forget the man!) still tricky even when you do it wrong :O
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3 months ago
#285

The_annoying_Devil

   3
2008-08-12 18:50:49, Chairman.Meow wrote:
#280 So is this a number written in base-10 that contains only 0's and 1's or is it a number written in binary?


Decimal, so base-10!
100 really means a hundred here an not 4.
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3 months ago
#286

Chairman.Meow

   3
2008-08-12 08:41:37, The_annoying_Devil wrote:
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:



Maybe I'm missing something here, but I don't think there's any rule regarding the sum of the digits in a square number (I listed the first 25 squares and the sum of their digits and there doesn't seem to be any pattern or any similarities). Can you give another hint?
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3 months ago
#287

Chairman.Meow

   3
2008-08-12 19:42:42, user_error wrote:
"#250 You're a jerk! Though the math isn't hard: 7^4."

Close, it would be 7^4 + 1 (don't forget the man!) still tricky even when you do it wrong :O


Actually it would be 7^4 + 2 (don't forget the narrator) ^^
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3 months ago
#288

The_annoying_Devil

   3
Actually its only one. "I was going to St Ives". The other people were already there. So only "I" went to St ives. 1!
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3 months ago
#289

Chairman.Meow

   3
2008-08-12 22:37:22, The_annoying_Devil wrote:
Actually its only one. "I was going to St Ives". The other people were already there. So only "I" went to St ives. 1!


Yes yes, we understand the actual answer to the riddle, we were talking about the math required when you do it wrong ^^. Stupid I guess.

What about my question a few posts ago?
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3 months ago
#290

The_annoying_Devil

   3
2008-08-12 22:27:05, Chairman.Meow wrote:
Maybe I'm missing something here, but I don't think there's any rule regarding the sum of the digits in a square number (I listed the first 25 squares and the sum of their digits and there doesn't seem to be any pattern or any similarities). Can you give another hint?


#hint2
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3 months ago
#291

The_annoying_Devil

   3
@Saline

Yep, its about eliminating certain numbers. The first one is, as you wrote, to eliminate those that have just two prim factors (and primes only of course). Then you can use Goldbach a little bit.

Found a second easier one similar to the first one (with the three ages). The only difference is that the sum of the ages is 13 and the product the "housenumber". And again one oldest child.
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3 months ago
#292

Chairman.Meow

   3
Devil, what does that have to do with the sum of the digits in Z^2? I have absolutely no idea where to go on this one and my head is starting to hurt. Can you just give a proof?
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3 months ago
#293

The_annoying_Devil

   3
Here it is:
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3 months ago
#294

Chairman.Meow

   3
AH! Immediately after my post I thought "divisibility test?", but didn't look any up. Nice one! Here's an easy one I just found.

How can you express the number 100 using six nines and no other digits?



Here's a similar one. My friend told me about a game he used to play in school when he was a kid, called "24". On a card you'd be given 4 numbers between 1 and 9 (inclusive). The goal of the game was to make 24 in as many ways as you can using those four numbers, addition, subtraction, multiplication, division, exponentiation, and parentheses. (Actually he never said anything about exponentiation but I imagine it's allowed.) After a certain time the teacher would say "stop" and the kid(s) with the most number of solutions would win.

As an example, given 2, 4, 5, and 8 you can do:

(4*5) + (8/2) = 24 and
((8*5)/2) + 4 = 24

We were drinking in the park one day throwing out sets of 4 numbers and playing the game, when we came across one that I still to this day can't solve: 2 7 8 9. Can anybody do it? Or prove that it's not possible?

I'm thinking of writing a computer program for this . . .

Edit: My program worked!!
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3 months ago
#295

Saline

   3
#294:
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3 months ago
#296

The_annoying_Devil

   3
Who has shot the ball against my car?
Arthur: "It was Bam."
Bam: "It was David."
Christian: "It was not me."
David: "Bam lied."

Who shot the ball...
a) if three boys tell the truth and one lies?
b) if one boy tells the truth and three lie?
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3 months ago
#297

Chairman.Meow

   3
2008-08-13 09:20:49, The_annoying_Devil wrote:
Who has shot the ball against my car?
Arthur: "It was Bam."
Bam: "It was David."
Christian: "It was not me."
David: "Bam lied."

Who shot the ball...
a) if three boys tell the truth and one lies?
b) if one boy tells the truth and three lie?
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3 months ago
#298

Chairman.Meow

   3
A Panda Bear walked into a resturant. He sat down at a table and ordered some food. When he was finished eating, he took out a gun and shot his waiter. He then left the resturant. After the police caught up with him, they asked him why he had killed the waiter. He replied, "Look me up in the dictionary." What did the dictionary say?
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3 months ago
#299

Chairman.Meow

   3
What is:

The beginning of eternity
The end of time and space
The beginning of every end
And the end of every place
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3 months ago
#300

fym.lpjuunin

   1
2008-08-13 20:39:18, Chairman.Meow wrote:
A Panda Bear walked into a resturant. He sat down at a table and ordered some food. When he was finished eating, he took out a gun and shot his waiter. He then left the resturant. After the police caught up with him, they asked him why he had killed the waiter. He replied, "Look me up in the dictionary." What did the dictionary say?




ROFL. That one almost killed me...^^
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