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

[fresh_42]

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]

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]

Spoiler

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]

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

Spoiler

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]

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]

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

 

[fresh_42]

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]

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]

Spoiler

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

 

[fresh_42]

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

 

[fresh_42]

Too complicated?

 

[lpetrich]

I have a partial solution:

Spoiler

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]

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]

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]

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]

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]

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 $\left( \dfrac{3}{p} \right) = -1$.

Huh?

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]

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]

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]

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]

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

 

[fresh_42]

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.

"No," answer the two boys.

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]

The teacher is surprised and looks again in her documents: both have the same birthday and the same parents.

Spoiler

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]

Spoiler

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.

 

[lpetrich]

Problem 71:

Spoiler

Let us consider one of the circles, and let us give it radius 1. The squares next to it in size are the inscribed square, the smaller one, and the circumscribed square, the larger one.

The inscribed square has its vertices on the circle. By symmetry, its center coincides with the center of the circle, O. Two neighboring vertices V1 and V2 have distances OV1 = OV2 = 1, and their angle is V1OV2 = 90d. By Pythagoras's theorem, distance V1V2 = sqrt(2).

The circumscribed square has the circle tangent to its edges. Consider two neighboring tangent points, T1 and T2. The tangent line for OT1 is perpendicular to it, and thus parallel to OT2. Likewise, the tangent line for OT2 is parallel to OT1. The intersection of those tangent lines I will call X. Angles T1OT2, XT1O, and OT2X are all 90d, making T2XT1 also 90d. This makes quadrangle T1OT2X a rectangle, and since OT1 and OT2 both have length 1, T1OT2X is a square with all sides having length 1. Since the tangent lines extend in both directions, the distance from one intersection to a neighboring one is thus 1+1 = 2.

Thus, the circumscribed square has side length 2, sqrt(2) times the inscribed square's side length. Taking the square gives the area, meaning that the circumscribed square has an area 2 times that of the inscribed square.

Thus, the ratio of the areas of two consecutive squares in this construction is 2.

 

[fresh_42]

Here is an easy way to see / solve it:

 

[mfb]

What is complicated about the adoption?

Is the year the same?

A mistake in the sheet the teacher got?

 

[fresh_42]

What is complicated about the adoption?

Adoption doesn't make them look alike.

Is the year the same?

Yes.

A mistake in the sheet the teacher got?

No.

 

[mfb]

Well, they could be half-brothers (shared father). Not technically an adoption I guess but similar.
They could have the same father and their biological mothers could be twins. That would make them genetically as similar as (non-identical) twins and it would be the result of questionable family planning choices.

"They lied" is a very simple possible answer, but I guess that is not the intended answer.

 

[fresh_42]

"They lied" is a very simple possible answer, but I guess that is not the intended answer.

Right. The real reason is completely natural and the easiest possible.

 

[fresh_42]

Let me phrase it with a mathematical anecdote my mentor once told me.
He listened to a dialogue between two students in the university's elevator.
One student was rather upset about the exam she just has had.
"This idiot asked me how often the constant function can be differentiated and I said once. This stupid question ruined a better grade."

So is the constant function exactly twice differentiable?

 

[DavidSnider]

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.

"No," answer the two boys.

The teacher is surprised and looks again in her documents: both have the same birthday and the same parents. How can that be?D88

Spoiler: spoiler

Surrogacy?

 

[fresh_42]

No. Or let's say: It could have happened in ancient Greece.

Edit: No Disney fans here?

 

[fresh_42]

73. $45$ must be divided in four parts, such that we get the same number if we add $2$ to the first part, subtract $2$ from the second, multiply the third by $2$ and divide the fourth by $2$.

D89

 

[pbuk]

#73

Spoiler

8, 12, 5, 20

 

[fresh_42]

74. Enter the numbers $1,2,\ldots,10$ into the circles, such that the sums of all numbers along the three inner triangles are equal.

D89

 

[mfb]

So is the constant function exactly twice differentiable?

Triplets (or more) then.

 

[lpetrich]

Proof of #73

Spoiler

In symbolic form, $45 = n_1 + n_2 + n_3 + n_4$ where $n_1 + 2 = n$, $n_2 - 2 = n$, $2n_3 = n$, and $n_4/2 = n$ for some number n.

One can easily find n1 to n4 from these: $n_1 = n - 2$, $n_2 = n + 2$, $n_3 = n/2$, $n_4 = 2n$. Substituting into the equation for 45, I find
$$ 45 = (n-2) + (n+2) + (n/2) + 2n = 9n/2 $$
Its solution is $n = 10$ and the four parts are $8, 12, 5, 20$.

 

[fresh_42]

75. Two trains passed each other in opposite directions, one at a speed of 36 km / h, the other at 45 km / h. A passenger on the second train counted and found, that the first train to pass him needed six seconds.
How long was the train?

D90

 

[mfb]

74. Enter the numbers $1,2,\ldots,10$ into the circles, such that the sums of all numbers along the three inner triangles are equal.

The center is part of the sum for all triangles, we can fill it last with whatever number is remaining and ignore it from now on. For every pair of triangles there are two numbers that are shared between these triangles, and one number that is exclusive to the other triangle. We can find a solution if we can find three mutually exclusive triples that satisfy a+b=c where a+b are put in the shared numbers and c is put on the opposite side. That way the sum in every triangle will be the sum of all numbers apart from the three outer numbers.
Doesn't need much searching to find 1+9=10, 2+6=8, 3+4=7.
The leftover number for the center is 5 and the sum in every triangle is 30, which means the average per field is 5 as well. Nice symmetry around 5 despite the asymmetric start (average number we fill in is 5.5).

 

[fresh_42]

View attachment 246474

Very interesting. I have a solution with 38 instead of 30:

I didn't think that more than one solution is possible. Maybe I should have asked how many sums are possible.

 

[lpetrich]

For #74, I wrote a short program that does a brute-force search. The number of permutations searched was 10! = 3628800, and I found 6528 solutions. Because none of the solutions have any symmetries, the total number of inequivalent solutions is the total number divided by the size of the triangle symmetry group (6): 1088.

I wrote it in C++, and I used STL <algorithm> function next_permutation() to iterate through the possibilities. The Standard Template Library is *great*.

 

[fresh_42]

For #74, I wrote a short program that does a brute-force search. The number of permutations searched was 10! = 3628800, and I found 6528 solutions. Because none of the solutions have any symmetries, the total number of inequivalent solutions is the total number divided by the size of the triangle symmetry group (6): 1088.

I wrote it in C++, and I used STL <algorithm> function next_permutation() to iterate through the possibilities. The Standard Template Library is *great*.

Have you calculated the possible sums in every solution?

 

[lpetrich]

Have you calculated the possible sums in every solution?

I have just done so. The number of solutions for each sum value is;
28: 96 16
29: 192 32
30: 480 80
31: 864 144
32: 1248 208
33: 768 128
34: 1248 208
35: 864 144
36: 480 80
37: 192 32
38: 96 16
with the total number and the number to within symmetries.

 

[mfb]

There is more symmetry. There is the factor 6 for the arrangement of the three triangles, but there is also a factor 8 from swapping numbers on the diagonals - different pattern but effectively the same solution. This is the reason all your numbers are divisible by 8. Removing this symmetry we end up with 136 options.

There is a more subtle symmetry: Replacing 1,2,...,10 by 10,9,...,1 leads to a solution as well (with triangle sum 66-x instead of x). That leads to the symmetric pattern of options you found. Removing this symmetry as well we end up with 68 options.

Is there a solution that also has the same sum for the big outer triangle?

 

[lpetrich]

There is more symmetry. There is the factor 6 for the arrangement of the three triangles, but there is also a factor 8 from swapping numbers on the diagonals - different pattern but effectively the same solution. This is the reason all your numbers are divisible by 8. Removing this symmetry we end up with 136 options.

How does that symmetry work? I find it difficult to picture.

 

[mfb]

Just swap e.g. 3 and 4 in the solution I posted. You can also swap 1 and 9 or 2 and 6.

 

[lpetrich]

So it's swap outer and middle along each diagonal.

Is there a solution that also has the same sum for the big outer triangle?

Indeed there is.

Inner-triangle length-equal solutions
28: 96 16 2
29: 192 32 4
30: 480 80 10
31: 864 144 18
32: 1248 208 26
33: 768 128 16
34: 1248 208 26
35: 864 144 18
36: 480 80 10
37: 192 32 4
38: 96 16 2
Total: 6528 1088 136
(raw, triangle symmetry, diagonal-interchange symmetry)

Equal to the outer-triangle length
31: 48 8
32: 120 20
34: 120 20
35: 48 8
Total: 336 56
(raw, triangle symmetry)

 

[fresh_42]

76. $n^3+ n^2 u +n^2 v+n^2w+nuv+nuw+nvw+uvw = 27,673,509,091$ with $u<v<w$.
What is $u\cdot n^2\,?$ (All numbers are non negative integers and $n$ maximal among all solutions.)

D91