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

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fresh_42

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67. "When you sit in the tram in the afternoon," says one man, "you get the impression that there are three times as many women as men." His neighbor quickly scanned the number of passengers and said, "That's true, at least for them Tram."

Now the tram stopped, and four times as many women got off as men got in. "Now the relationship is a bit more bearable. There are only twice as many women as men in here," the man said to his neighbor.

At the next stop there was only one woman. Since no one was about to get out, the man said to his neighbor, "Come on, we're going out of here, then the old ratio of 1: 3 is restored."

How many people continued on the train after the men left and the lady got in?

D80

mfb

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Interesting general strategy with the divisor puzzle. I expected this to be one of the puzzles where a general solution can be complicated so after I found a proof that there is a strategy (but not the strategy) I didn't look much further.

lpetrich

I will express the number of passengers as (number of men, number of women).

Initially, there are $(m_1, 3m_1)$ passengers. At the first stop, $(m_2,-4m_2)$ arrive, since departure is negative arrival. This leaves $(m_3, 2m_3)$.

This part has the solution $m_1 = 6n, m_2 = n, m_3 = 7n$, giving $(6n,18n)$, $(n,-4n)$, $(7n,14n)$.

At the second stop, the arrivals and departures are $(-2,1)$, and the result is $m_4, 3m_4$. Solving gives $n = 1, m_4 = 5$.

Initially, the tram has (6,18). At the first stop, (1,-4) arrive, giving (7,14). At the second stop, (-2,1) arrive, giving (5,15). Thus, a total of 20 people continuing on the tram after its second stop.

fresh_42

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68. A pond owner has an unknown amount of fish in his pond. Since he wants to count them now, he catches 30 fish, marks them with red paint and throws them back into the pond. The next day he catches 40 fish. Two of these fish have a red mark.

How many fish are approximately in the pond?

D82

lpetrich

#68
This is a classic method of wildlife census-taking.

Let us say that there are $N$ fish in the pond. After the first fish catching, there are now $N_m$ marked fish in it. The second time around, the pond owner caught $N_2$ fish, of which $N_{m2}$ are marked. The probability of a fish being marked is $p_m = N_m/N$, and the most likely number of marked fish caught the second time around is $N_{m2} = N_2 p_m = \frac{N_2 N_m}{N}$. This gives us $N = \frac{N_2 N_m}{N_{m2}}$.

Working out the numbers, $N = \frac{40 \cdot 30}{2} = 600$. So the pond owner has 600 fish.

mfb

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If you do this one fish at a time it has an interesting relation to the birthday problem. Consider a pond with 365 fish in it, fish one at a time at random, mark it. When does the first marked fish appear? This is equivalent to drawing random people from a large population, marking their birthdays, and looking for the first duplicate. The answer is well-known: 23 for a ~50% chance[/url]. If we find the first duplicate after 23 fish then we can use 365 as estimate for the total number of fish. It won't be a good estimate with just a single match, of course, but if we look for the second, third, ... we can get better estimates. With the same number of fish you have more information than the two-batch approach.

BvU

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plus or minus how many ? What's the proper standard deviation estimate ? (in #305)

fresh_42

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plus or minus how many ? What's the proper standard deviation estimate ? (in #305)
Well, this is not the thread to ask for $P(590 \leq X \leq 610) \geq .95$.

fresh_42

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69. What time is it if so many minutes are missing until 9 o'clock, as 40 minutes ago 3 times as many minutes after 6 o'clock had elapsed?

D82

BvU

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8:25
from:180 - x = 3x + 40 => x = 35

fresh_42

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This is an experiment with a self made puzzle. So try not to complain too much.

70. I am the product of two odd primes, and can be written only using $2,3$ and at most two $1$. This representation is asked for. My primes are two out of the first three consecutive prime sequence members of the first (minimal first element) sequence with this property, defined as: the ratio of the predecessors in $\mathbb{N}$ of two consecutive sequence elements is $2$. Btw., for my primes $p$ we have $\left( \dfrac{3}{p} \right) \neq -1\,.$

D82

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Too complicated?

lpetrich

I have a partial solution:
For decimal digits (0 to 2) of 1, (1) of 2, and (1) of 3 and divisibility by exactly two odd primes, I find three numbers:
• 123 = 3*41
• 213 = 3*71
• 321 = 3*107
But beyond this, I am totally stumped.

fresh_42

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You could solve the recursion and test which values $a_0$ lead to three consecutive primes (in the first four sequence elements.) I guess I should have added this, for otherwise there are too many sequences possible.
Excel or a similar program should do the job.

mfb

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Too confusing.
the ratio of the predecessors in N of two consecutive sequence elements is 2
If p,q are elements then (p-1)/(q-1)=2 or equivalently p=2q-1, okay.
My primes are two out of the first three consecutive prime sequence members of the first (minimal first element) sequence with this property
We look for the sequence with the minimal first element. 2->3->5 is the first set of primes that satisfy the condition above, so it should be the first sequence. As 9 is not prime the sequence ends at 5. 2*3, 2*5, 3*5 don't satisfy the condition on the digits of the product. What am I missing?
$\left( \dfrac{3}{p} \right) \neq -1\,.$
Clearly this is not just a regular fraction. I think I have seen that notation in the context of divisors before but don't remember what it was. It would help to clarify the notation (searching for the usual keywords is completely pointless here). I also don't know why this would be relevant.

fresh_42

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No, we look for any three consecutive primes among $\{\,a_0,\ldots,a_4\,\}$, and odd ones. I thought the sequence is easy.

Btw., is DESY still in operation?

mfb

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No, we look for any three consecutive primes among $\{\,a_0,\ldots,a_4\,\}$, and odd ones. I thought the sequence is easy.
Okay, let's look further.
For starting primes: They can't end in 3 as that would make the second number divisible by 5 (and 3,5,9 doesn't have a prime as third number) and they can't end in 7 as that makes the third number divisible by 5. Divided by 3 they must have remainder 1.

The next sequence is 19,37,73, but these products don't work either. After that there are 79,157,313 and 109,217,433, same issue. The starting prime must be at least 200.
Btw., is DESY still in operation?
Huh?

fresh_42

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The next sequence is 19,37,73, but these products don't work either.
In which sense? They are the first primes with the required property in the first sequence. $a_0 \in \{\,0,1,\ldots,9\,\}$ do not work, $a_0=10$ does. Now one of it has $\left( \dfrac{3}{p} \right) = -1$.
I thought I could as well ask you instead of searching the web. The DESY site didn't answer the question at once, so I would have had to start to read it.

As I answered another thread I realized, that I know the machines in Darmstadt and Karlsruhe, but wasn't sure whether Hamburg and Adlershorst are still in operation for research and studies. As it wasn't really important, I thought I simply ask you.

mfb

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19*37=703, 18*73=1387, 37*73=2701. They all have a 7 and a 0 or 8. We could look at the numbers in base 4 where 703_10=22333_4 but I think this is not the spirit of the question.
Now one of it has $\left( \dfrac{3}{p} \right) = -1$.
You still didn't help with that notation.
I thought I could as well ask you instead of searching the web. The DESY site didn't answer the question at once, so I would have had to start to read it.
Quite off-topic here. DESY still runs, mainly as preaccelerator for PETRA which is used for photon science.

fresh_42

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It's the Legendre symbol which measures the quadratic residue.
$a$ is a quadratic residue modulo $m$ if $(a,m)=1$ and there is an $x$ such that $a\equiv x^2 \mod m$.
$$\left( \dfrac{3}{p} \right) = \begin{cases} 1 \quad &\text{ if } 3 \not\equiv 0 \mod p \text{ is a quadratic residue }\mod p \\ 0 \quad &\text{ if } 3 \equiv 0 \mod p \\ -1 \quad &\text{ if } 3 \text{ is a non-quadratic residue }\mod p\end{cases}$$
Legendre defined it via the Euler criterion $\left( \dfrac{3}{p} \right) = 3^{\frac{p-1}{2}}$ for odd primes.

'Can be written as' does not mean 'can be factorized with'. Given the sequence formula, the presentation with $1,2,3$ drops almost automatically into place. Finding the sequence is the only really hard part.

mfb

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Legendre symbol, thanks.
'Can be written as' does not mean 'can be factorized with'. Given the sequence formula, the presentation with $1,2,3$ drops almost automatically into place. Finding the sequence is the only really hard part.
If we can define the notation then every number can be written like this.
The question is still too confusing for me.

fresh_42

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Let's forget it. I was looking for $(2^2\cdot 3^2 +1)\cdot (2^3\cdot 3^2 +1)$.

71. If we build a geometric Matryoshka doll with an inscribed circle in an inscribed square in an inscribed circle in an inscribed square ... in an inscribed circle in a square. What are the ratios of the areas of two consecutive squares?

D85

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fresh_42

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72. Summer is over and there are two new students in the class who are hard to tell apart. "You are certainly twins?", asks the teacher.

The teacher is surprised and looks again in her documents: both have the same birthday and the same parents. How can that be?

D88

jbriggs444

Homework Helper
The teacher is surprised and looks again in her documents: both have the same birthday and the same parents.
Skipping the possibility that the students are adopted, let us go for the idea that they are similar looking siblings born one year apart.

Possibly born close to the cutoff for a year's batch of children to enter the first year of school, straddling a year when the eligibility dates where shifted. Possibly one child held back or skipping a grade. Or possibly in college, one in first year and one in second year, both attending the same class.

Possibly born to a lesbian couple.

fresh_42

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Skipping the possibility that the students are adopted, let us go for the idea that they are similar looking siblings born one year apart.

Possibly born close to the cutoff for a year's batch of children to enter the first year of school, straddling a year when the eligibility dates where shifted. Possibly one child held back or skipping a grade. Or possibly in college, one in first year and one in second year, both attending the same class.

Possibly born to a lesbian couple.
There is a simpler explanation.

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

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