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

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#### fresh_42

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7. Jennifer likes to drink black tea. One day, a friend tells her that tea tastes very good with a few drops of freshly squeezed lemon. Jennifer decides to test it the next morning. As the water starts to boil, she squeezes out a lemon and puts a teabag in the cup.

Jennifer has the following information as a math-ace in her class:
- The cup has a capacity of 10 cl.
- The tea bag has a volume of 1 cl.
- She has 1 cl of lemon, which she wants to give in the tea (temperature: 20 ° C).
- She has 8 cl of hot water (temperature: 100 ° C).
- She knows she needs exactly 5 minutes to make her school sandwiches.

Jennifer now faces the following question:

When does my tea get cold faster? If I put the lemon in the cup right away, or if I do that after I made the school sandwiches?

#### fresh_42

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Come on, folks, nobody? This is a physics website!

#### jbriggs444

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Come on, folks, nobody? This is a physics website!
It's a mathematics forum in a physics website. There is no mathematical solution because the problem is not completely specified.

Bringing physics into the situation... Primary heat loss in 100 degree hot water is likely to be evaporation. Early mixing to bring that temperature down is going to be a win.

#### fresh_42

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Early mixing to bring that temperature down is going to be a win.
Do you mean a win in the sense that early mixing results in a cooler tea?

#### jbriggs444

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Do you mean a win in the sense that early mixing results in a cooler tea?
Rather, early mixing would lead to a warmer tea since the net heat loss rate is lowered.

I was assuming that the goal was warmer tea, not colder. But perhaps the opposite is desired.

#### fresh_42

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When does my tea get cold faster?
Rather, early mixing would lead to a warmer tea since the net heat loss rate is lowered.
Correct. Early mixing lowers the temperature difference of tea and air, so the tea will not cool down as fast as if it was when boiling. So a late mixing is cooler.

#### fresh_42

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8. This sequence is arranged according to which rule?
$$8\quad 5\quad 4 \quad 9 \quad 1\quad 7 \quad 6 \quad 3 \quad 2 \quad 0$$

#### Orodruin

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Correct. Early mixing lowers the temperature difference of tea and air, so the tea will not cool down as fast as if it was when boiling. So a late mixing is cooler.
I think I agree with @jbriggs444 that this is not really a math problem. As specified, it requires a non-zero amount of physical modelling to have a solution.

8. This sequence is arranged according to which rule?
$$8\quad 5\quad 4 \quad 9 \quad 1\quad 7 \quad 6 \quad 3 \quad 2 \quad 0$$
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.

mfb

#### fresh_42

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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.
Sure. So find one. Of course there is an easy solution, but maybe we get some funny complicated stuff here. And please do not use $\prod_k (x-k)^{n_k}=0$ or similar trivial but high order solutions.

#### lpetrich

8. This sequence is arranged according to which rule?
$$8\quad 5\quad 4 \quad 9 \quad 1\quad 7 \quad 6 \quad 3 \quad 2 \quad 0$$
The English names of these numbers in alphabetical order:

eight, five, four, nine, one, seven, six, three, two, zero

#### fresh_42

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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.

#### fresh_42

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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?

#### mfb

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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.

#### lpetrich

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.

#### mfb

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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.

#### fresh_42

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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.

#### fresh_42

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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.

#### fresh_42

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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?

#### Orodruin

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Tim is riding Jim’s horse.

#### fresh_42

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11.

How did Carlsen (B) to draw win and why?

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#### mfb

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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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#### fresh_42

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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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#### fresh_42

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12.

In which order are the cups filled with coffee?

#### lpetrich

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.

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

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