Challenge Riddles and Puzzles: Extend the following to a valid equation

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The discussion revolves around extending mathematical expressions to create valid equations, with participants sharing various solutions for sequences like 9 9 9 = 6 and 8 8 8 = 6. A secondary topic involves a game played on a circular field where players place coins, prompting questions about the fairness of the game based on the ratio of the field's radius to the coin's radius. Participants express confusion over the concept of fairness in deterministic games that cannot end in a draw, leading to deeper discussions about game theory. Additionally, there are puzzles involving number functions and urns with colored balls, showcasing the group's engagement with logical reasoning and problem-solving. The conversation highlights the blend of mathematical challenges and playful banter among participants.
  • #61
9. We are looking for a pair of numbers. Mr. P knows the product of them and Mr. S their sum. The numbers are one of the pairs: ##(3,5)\, , \,(8,11)\, , \,(2,7)\, , \,(4,13)##. We heard the following dialogue of the two:
  • Mr. P: I don't know the numbers.
  • Mr. S: I don't know them either, but I knew you couldn't know them.
  • Mr. P: Then I know the two numbers by now.
  • Mr. S: Me, too.
 
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  • #62
Orodruin said:
We cannot really say. There could be many valid rules that would produce that exact sequence. Which one of those you used is not possible to determine.
As I was searching for new puzzles I stumbled upon one which was quite easy to solve, i.e. it took me only a few seconds and was too easy to qualify for here. It was in a newspaper and they titled: Who can solve this has an IQ of 150.

My first reaction was: never ever. Now I think, what if they are right? 100 is said to be the average. What does this tell us about the state of our societies?
 
  • #63
These newspapers don’t want to tell their readers they are stupid - not even the stupid readers. Some readers will try and give up. If you tell them “everyone with average intelligence should be able to solve this” you might lose some readers. If you tell them “you need [high number] IQ then they don’t feel stupid.
 
  • #64
fresh_42 said:
9. We are looking for a pair of numbers. Mr. P knows the product of them and Mr. S their sum. The numbers are one of the pairs: ##(3,5)\, , \,(8,11)\, , \,(2,7)\, , \,(4,13)##. We heard the following dialogue of the two:
  • Mr. P: I don't know the numbers.
  • Mr. S: I don't know them either, but I knew you couldn't know them.
  • Mr. P: Then I know the two numbers by now.
  • Mr. S: Me, too.
Some big problems are that the sums and products both have ambiguous decompositions, and also that none of these number pairs' sums and products overlap -- they are all distinct.
 
  • #65
I assume P and S don't know that these four are the only options, otherwise P should know the numbers (the products are 15, 88, 14, 52, all unique).

P can know the numbers if and only if the product P is a prime, otherwise (1,P) and (n,P/n) are options for a non-trivial divisor n. None of the products is a prime, so the first line doesn't help us. Let's look at S' reply:
If the sum is 8 then the product could be 7 (1,7) and P could know the numbers. S knows this cannot be the case, therefore the sum cannot be 8 and the numbers can't be (3,5).
If the sum is 19 then the numbers are (n,19-n) and n(19-n) is not a prime for any n (for n=1 we get 18). Similarly for 9 and 17 as sum. In all three cases S knows P cannot know the numbers. In fact p+1 for some prime p is the only sum where the product might be unique: For all other numbers S knows P cannot know the numbers.

Let's go back to P.
What could the numbers be if the product was 88? (1,88), (2,44), (4,22), (8,11) are all options, leading to sums 89, 46, 26, 19. None of these is one larger than a prime. If the product is 88 then P knows from the beginning on that S knows that they cannot determine the numbers.
What could the numbers be if the product was 14? (1,14) and (2,7) are the only options, leading to sums 15 and 9. Again none of these is one larger than a prime.
What could the numbers be if the product was 52? We have (1,52), (2,26), (4,13) with sums 53,28,17. Same here.
All three options looks possible. I don't see how P could continue to rule out some of them based on what S said.
 
  • #66
lpetrich said:
Some big problems are that the sums and products both have ambiguous decompositions, and also that none of these number pairs' sums and products overlap -- they are all distinct.
Assume they don't know the split.
 
  • #67
The solution to number 9. I was looking for was the pair ##(4,13)##. The clue was to rule out all possibilities of the form ##pq## which allowed a unique factorization by comparison to the according level of information.
 
  • #68
10. Once upon a time there were two good friends, Tim and Jim. One day, they came up with the idea of making a small bet. It was supposed to be a small horse race organized by John. He set the rules as follows:

The race starts at the church, from there the two horses have to cross a hill, past a bus stop, then along the lake and finally back to the church! Of course the way had to be kept exactly! In addition, whose horse is the last to arrive again at the church, whose master is the winner and has won the bet!

But after John arranged and organized everything, something very strange happened. At first Tim and Jim looked at each other questioningly and wondered how they should proceed best, but then Tim suddenly stormed off, ran like a wolf, jumped on the horse, swept off and rode over the hill past the bus stop, along the lake and back to the church. Jim stood there and did not know what happened to him, started way too late and could not catch up.

John watched the race and when Tim arrived at the finish, he congratulated him and said that he had won the race!

How could this happen?
 
  • #69
Tim is riding Jim’s horse.
 
  • #70
11.
242560

How did Carlsen (B) to draw win and why?
 
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  • #71
fresh_42 said:
The solution to number 9. I was looking for was the pair ##(4,13)##. The clue was to rule out all possibilities of the form ##pq## which allowed a unique factorization by comparison to the according level of information.
What was the error in my reasoning then?
pq does not allow a unique factorization as (1,pq) is a valid pair of numbers. You didn't specify that 1 cannot be a number.
 
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  • #72
mfb said:
What was the error in my reasoning then?
pq does not allow a unique factorization as (1,pq) is a valid pair of numbers. You didn't specify that 1 cannot be a number.
Yes, you are right. I had forgotten to mention that the numbers are properly between 1 and 100.
I don't like chess puzzles that don't specify who is who and who moves next.
Well, I did specify it:
  • Meier vs. Carlsen (This implies Meier white and Carlsen black.)
  • How did Carlsen win? (This implies black to draw.)
 
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  • #73
12.
242589


In which order are the cups filled with coffee?
 
  • #74
My solution:
The coffee is poured into the top chamber, and as that chamber is filled, the coffee level reaches the pipe on the left.

When it does so, it pours into the left chamber. As it does, it reaches the left pipe from it. That pipe is blocked at the chamber, so the coffee continues to fill that chamber. Cup 4 thus never receives any coffee. When it gets to the level of the right pipe from that chamber, it flows into that pipe, but it is blocked at the spout end. Cup 9 never receives any coffee, and the coffee continues to fill the left chamber, filling it and the pipe leading into it.

It fills the top chamber until it reaches the pipe on the right, and it then pours into the top right chamber, and then the bottom right one. When it fills the bottom right one, it starts filling the top right one until it meets the pipe on the right. The coffee pours into that pipe but is blocked at the joint. Cup 7 never receives any coffee.

The top right chamber continues to become filled until the coffee level reaches the pipe to the left, and it then pours into that pipe. It is unobstructed, and the coffee pours into Cup 5.

Thus, cup 5 is the only cup of these four that ever receive any coffee.
 
  • #76
fresh_42 said:
No chess fans here?
https://www.physicsforums.com/threa...o-a-valid-equation.970213/page-3#post-6171170
I'm a bit disappointed.
For what it's worth, I have beaten lvl-1 stockfish (on here) a couple of times (without any move-backtracking) ... and losing many times :P.

I actually looked at it for quite some time yesterday. I just couldn't find a way to find a quick victory. Eventually, I searched-up and just looked at the game, but well ... I am still not sure. I thought of that particular move (for black). The response of white that I couldn't resolve isn't taken in actual game (likely for good reasons, but somehow I don't get it). But maybe if I took up pen and paper and tried to sketch every possibility, I could be able to do it.

But since I already watched, it would be better for me not to post anything.
 
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  • #77
SSequence said:
But since I already watched, it would be better for me not to post anything.
For one there is the thread to run into a position to lose the queen by chess through the knight, and the other option is to run away with the king which ultimately loses a knight or allows the pawn to convert.
 
  • #78
I just forgot about pawn promotion completely. It should likely be solveable with promotion.
 
  • #80
DavidSnider said:
Natural numbers?
Yes, and with a decimal length less than ten.
 
  • #81
10^9 - 9^9 = 612579511
 
  • #82
DavidSnider said:
10^9 - 9^9 = 612579511
Correct, but you could have said how you did it. Not because it is difficult, or can be guessed by the formula, but because it is an important technique in counting and often saves time, programming or even an induction:
It is often easier to count the complement!
 
  • #83
15. What is the smallest prime which includes all ten digits exactly once?
 
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  • #84
I have not seen the word 'cipher' used that way before in English - I think most people would ask "What is the smallest prime which includes all ten decimal digits exactly once".

The word 'digit' probably makes it easier to get to the root of the problem.
 
  • #85
pbuk said:
I have not seen the word 'cipher' used that way before in English - I think most people would ask "What is the smallest prime which includes all ten decimal digits exactly once".

The word 'digit' probably makes it easier to get to the root of the problem.
Thanks, corrected. The difficulty is to distinguish between digit as a number and digit as a position, cause you also say: estimate up to three digits, which would be position. As I meant the symbol, and neither value nor position, I thought cipher would be adequate.
 
  • #86
fresh_42 said:
15. What is the smallest prime which includes all ten digits exactly once?
I seem to have killed this, so I suppose I had better answer it and we can move on!
There is no such prime. The sum of the digits 0...9 is 45 which is divisible by 9, and so any integer which includes each of these digits exactly once is also divisible by 9.
 
  • #87
16. A man drove his car one mile to the top of a mountain at the rate of fifteen miles per hour. How fast must he drive one mile down the other side in order to average thirty miles per hour for the hole trip of two miles?
 
  • #88
fresh_42 said:
16. A man drove his car one mile to the top of a mountain at the rate of fifteen miles per hour. How fast must he drive one mile down the other side in order to average thirty miles per hour for the hole trip of two miles?
It's impossible. It takes 4 minutes to go 2 miles @ 30 MPH but the first half of the trip already took 4 minutes.
 
  • #89
17. You have a glass of 200 ml of water and one with 200 ml of wine. Now pour some water into the wine and after mixing, pour back so much again, until the water glass has again 200ml. 90% are still water. How big is the percentage of wine in the wine glass?
 
  • #90
fresh_42 said:
Now pour some water into the wine
This must be against some law, but ok ...

The water glass now contains 90% (180 ml) of water and 10% (20 ml) of wine. Thus, the wine glass must contain the remainder, i.e., 180 ml of wine (90%) and 20 ml of water (10%).
 

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